Strength Analysis in Geomechanics

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Strength Analysis inGeomechanics

Serguey A. Elsoufiev

With 158 Figures and 11 Tables

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V.I. BabitskyDepartment of Mechanical EngineeringLoughborough University

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J. WittenburgTechnische MechanikInstitut

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Serguey A. Elsoufiev

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Foreword

It is hardly possible to find a single rheological law for all the soils. However,they have mechanical properties (elasticity, plasticity, creep, damage, etc.)that are met in some special sciences, and basic equations of these disciplinescan be applied to earth structures. This way is taken in this book. It representsthe results that can be used as a base for computations in many fields of theGeomechanics in its wide sense. Deformation and fracture of many objectsinclude a row of important effects that must be taken into account. Some ofthem can be considered in the rheological law that, however, must be simpleenough to solve the problems for real objects.

On the base of experiments and some theoretical investigations the con-stitutive equations that take into account large strains, a non-linear unsteadycreep, an influence of a stress state type, an initial anisotropy and a damageare introduced. The test results show that they can be used first of all tofinding ultimate state of structures – for a wide variety of monotonous load-ings when equivalent strain does not diminish, and include some interrupted,step-wise and even cycling changes of stresses. When the influence of timeis negligible the basic expressions become the constitutive equations of theplasticity theory generalized here. At limit values of the exponent of a hard-ening law the last ones give the Hooke’s and the Prandtl’s diagrams. Togetherwith the basic relations of continuum mechanics they are used to describe thedeformation of many objects. Any of its stage can be taken as maximumallowable one but it is more convenient to predict a failure according to thecriterion of infinite strains rate at the beginning of unstable deformation. Themethod reveals the influence of the form and dimensions of the structure onits ultimate state that are not considered by classical approaches.

Certainly it is hardly possible to solve any real problem without someassumptions of geometrical type. Here the tasks are distinguished as anti-plane (longitudinal shear), plane and axisymmetric problems. This allowsto consider a fracture of many real structures. The results are representedby relations that can be applied directly and a computer is used (if neces-sary) on a final stage of calculations. The method can be realized not only in

VI Foreword

Geomechanics but also in other branches of industry and science. The wholeapproach takes into account five types of non-linearity (three physical andtwo geometrical) and contains some new ideas, for example, the considerationof the fracture as a process, the difference between the body and the elementof a material which only deforms and fails because it is in a structure, thesimplicity of some non-linear computations against linear ones (ideal plastic-ity versus the Hooke’s law, unsteady creep instead of a steady one, etc.), theindependence of maximum critical strain for brittle materials on the types ofstructure and stress state, an advantage of deformation theories before flowones and others.

All this does not deny the classical methods that are also used in the bookwhich is addressed to students, scientists and engineers who are busy withstrength problems.

Preface

The solution of complex problems of strength in many branches of industryand science is impossible without a knowledge of fracture processes. Last 50years demonstrated a great interest to these problems that was stimulatedby their immense practical importance. Exact methods of solution aimed atfinding fields of stresses and strains based on theories of elasticity, plastic-ity, creep, etc. and a rough appreciation of strength provide different resultsand this discrepancy can be explained by the fact that the fracture is a com-plex problem at the intersection of physics of solids, mechanics of media andmaterial sciences. Real materials contain many defects of different form anddimensions beginning from submicroscopic ones to big pores and main cracks.Because of that the use of physical theories for a quantitative appreciation ofreal structures can be considered by us as of little perspective. For technicalapplications the concept of fracture in terms of methods of continuum me-chanics plays an important role. We shall distinguish between the strength ofa material (considered as an element of it – a cube, for example) and that ofstructures, which include also samples (of a material) of a different kind. Weshall also distinguish between various types of fracture: ductile (plastic at bigresidual strains), brittle (at small changes of a bodies’ dimensions) and dueto a development of main cracks (splits).

Here we will not use the usual approach to strength computation when thedistribution of stresses are found by methods of continuum mechanics and thenhypotheses of strength are applied to the most dangerous points. Instead, weconsider the fracture as a process developing in time according to constitutiveequations taking into account large strains of unsteady creep and damage(development of internal defects). Any stage of the structures’ deformationcan be supposed as a dangerous one and hence the condition of maximumallowable strains can be used. But more convenient is the application of acriterion of an infinite strain rate at the moment of beginning of unstabledeformation. This approach gives critical strains and the time in a naturalway. When the influence of the latter is small, ultimate loads may be also

VIII Preface

found. Now we show how this idea is applied to structures made of differentmaterials, mainly soils.

The first (introductory) chapter begins with a description of the role ofengineering geological investigations. It is underlined that foundations shouldnot be considered separately from structures. Then the components of geo-mechanics are listed as well as the main tasks of the Soil Mechanics. Its shorthistory is given. The description of soil properties by methods of mechanics isrepresented. The idea is introduced that the failure of a structure is a process,the study of which can describe its final stage. Among the examples of thisare: stability of a ring under external pressure and of a bar under compressionand torsion (they are represented as particular cases of the common approachto the stability of bars); elementary theory of crack propagation; the ultimatestate of structures made of ideal plastic materials; the simplest theory ofretaining wall; a long-time strength according to a criterion of infinite elonga-tions and that of their rate. The properties of introduced non-linear equationsfor unsteady creep with damage as well as a method of determination of creepand fracture parameters from tests in tension, compression and bending aregiven (as particular cases of an eccentric compression of a bar).

In order to apply the methods of Chap. 1 to real objects we must intro-duce main equations for a complex stress state that is made in Chap. 2. Thestresses and stress tensor are introduced. They are linked by three equilib-rium equations and hence the problem is statically indeterminate. To solvethe task, displacements and strains are introduced. The latters are linked bycompatibility equations. The consideration of rheological laws begins with theHooke’s equations and their generalization for non-linear steady and unsteadycreep is given. The last option includes a damage parameter. Then basic ex-pressions for anisotropic materials are considered. The case of transversallyisotropic plate is described in detail. It is shown that the great influence ofanisotropy on rheology of the body in three options of isotropy and loadingplanes interposition takes place. Since the problem for general case cannotbe solved even for simple bodies some geometrical hypotheses are introduced.For anti-plane deformation we have five equations for five unknowns and thetask can be solved easily. The transition to polar coordinates is given. For aplane problem we have eight equations for eight unknowns. A very useful andunknown from the literature combination of static equations is received. Thebasic expressions for axi-symmetric problem are given. For spherical coordi-nates a useful combination of equilibrium laws is also derived.

It is not possible to give all the elastic solutions of geo-technical problems.They are widely represented in the literature. But some of them are includedin Chap. 3 for an understanding of further non-linear results. We begin withlongitudinal shear which, due to the use of complex variables, opens the wayto solution of similar plane tasks. The convenience of the approach is basedon the opportunity to apply a conformal transformation when the results forsimple figures (circle or semi-plane) can be applied to compound sections.The displacement of a strip, deformation of a massif with a circular hole and

Preface IX

a brittle rupture of a body with a crack are considered. The plane deformationof a wedge under an one-sided load, concentrated force in its apex and pressedby inclined plates is also studied. The use of complex variables is demonstratedon the task of compression of a massif with a circular hole. General relationsfor a semi-plane under a vertical load are applied to the cases of the crackin tension and a constant displacement under a punch. In a similar way rela-tions for transversal shear are used, and critical stresses are found. Among theaxi-symmetric problems a sphere, cylinder and cone under internal and exter-nal pressures are investigated. The generalization of the Boussinesq’s problemincludes determination of stresses and displacements under loads uniformlydistributed in a circle and rectangle. Some approximate approaches for a com-putation of the settling are also considered. Among them the layer-by-layersummation and with the help of the so-called equivalent layer. Short infor-mation on bending of thin plates and their ultimate state is described. As aconclusion, relations for displacements and stresses caused by a circular crackin tension are given.

Many materials demonstrate at loading a yielding part of the stress-straindiagram and their ultimate state can be found according to the Prandtl’s andthe Coulomb’s laws which are considered in Chap. 4, devoted to the ultimatestate of elastic-plastic structures. The investigations and natural observationsshow that the method can be also applied to brittle fracture. This approach issimpler than the consequent elastic one, and many problems can be solved onthe basis of static equations and the yielding condition, for example, the tor-sion problem, which is used for the determination of a shear strength of manymaterials including soils. The rigorous solutions for the problems of cracks andplastic zones near punch edges at longitudinal shear are given. Elastic-plasticdeformation and failure of a slope under vertical loads are studied amongthe plane problems. The rigorous solution of a massif compressed by inclinedplates for particular cases of soil pressure on a retaining wall and flow of theearth between two foundations is given. Engineering relations for wedge pen-etration and a load-bearing capacity of a piles sheet are also presented. Theintroduced theory of slip lines opens the way to finding the ultimate state ofstructures by a construction of plastic fields. The investigated penetration ofthe wedge gives in a particular case the ultimate load for punch pressure in amedium and that with a crack in tension. A similar procedure for soils is re-duced to ultimate state of a slope and the second critical load on foundation.Interaction of a soil with a retaining wall, stability of footings and differentmethods of slope stability appreciation are also given. The ultimate state ofthick-walled structures under internal and external pressures and compressionof a cylinder by rough plates are considered among axi-symmetric problems.A solution to a problem of flow of a material within a cone, its penetration ina soil and load-bearing capacity of a circular pile are of a high practical value.

Many materials demonstrate a non-linear stress-strain behaviour from thebeginning of a loading, which is accompanied as a rule by creep and damage.This case is studied in Chap. 5 devoted to the ultimate state of structures

X Preface

at small non-linear strains. The rigorous solution for propagation of a crackand plastic zones near punch edges at anti-plane deformation is given. Thegeneralization of the Flamant’s results and the analysis of them are presented.Deformation and fracture of a slope under vertical loads are considered interms of simple engineering relations. The problem of a wedge pressed byinclined plates and a flow of a material between them as well as penetrationof a wedge and load-bearing capacity of piles sheet are also discussed. Theproblem of the propagation of a crack and plastic zones near punch edges attension and compression as well as at transversal shear are also studied. Aload-bearing capacity of sliding supports is investigated. A generalization ofthe Boussinesq’s problem and its practical analysis are fulfilled. The flow of thematerial within a cone, its penetration in a massif and the load-bearing capa-city of a circular pile are studied. As a conclusion the fracture of thick-walledelements (an axi-symmetrically stretched plate with a hole, sphere, cylinderand cone under internal and external pressures) are investigated. The resultsof these solutions can be used to predict failure of the voids of different formand dimensions in soil.

In the first part of Chap. 6, devoted to the ultimate state of structuresat finite strains, the Hoff’s method of infinite elongations at the moment offracture is used. A plate and a bar at tension under hydrostatic pressure areconsidered. Thick-walled elements (axi-symmetrically stretched plate with ahole, sphere, cylinder and cone under internal and external pressure) are stud-ied in the same way. The reference to other structures is made. The secondpart of the chapter is devoted to mixed fracture at unsteady creep. The sameproblems from its first part are investigated and the comparison with theresults by the Hoff’s method is made. The ultimate state of shells (a cylinderand a torus of revolution) under internal pressure as well as different mem-branes under hydrostatic loading is studied. The comparison with test data isgiven. The same is made for a short bar in tension and compression. In con-clusion the fracture of an anisotropic plate in biaxial tension is investigated.The results are important not only for similar structures but also for a find-ing the theoretical ultimate state of a material element (a cube), which areformulated according to the strength hypotheses. The found independence ofcritical maximum strain for brittle materials on the form of a structure andthe stress state type can be formulated as a “law of nature”.

Contents

1 Introduction: Main Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Role of Engineering Geological Investigations . . . . . . . . . . . . . . . 11.2 Scope and Aim of the Subject: Short History

of Soil Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Use of the Continuum Mechanics Methods . . . . . . . . . . . . . . . . . . 21.4 Main Properties of Soils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4.1 Stresses in Soil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.2 Settling of Soil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4.3 Computation of Settling Changing in Time . . . . . . . . . . . 11

1.5 Description of Properties of Soils and Other Materialsby Methods of Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5.2 The Use of the Elasticity Theory . . . . . . . . . . . . . . . . . . . . 141.5.3 The Bases of Ultimate Plastic State Theory . . . . . . . . . . 171.5.4 Simplest Theories of Retaining Walls . . . . . . . . . . . . . . . . 211.5.5 Long-Time Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.5.6 Eccentric Compression and Determination of Creep

Parameters from Bending Tests . . . . . . . . . . . . . . . . . . . . . 26

2 Main Equations in Media Mechanics . . . . . . . . . . . . . . . . . . . . . . . 312.1 Stresses in Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.2 Displacements and Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.3 Rheological Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.3.1 Generalised Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3.2 Non-Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.3.3 Constitutive Equations for Anisotropic Materials . . . . . . 36

2.4 Solution Methods of Mechanical Problems . . . . . . . . . . . . . . . . . . 402.4.1 Common Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.4.2 Basic Equations for Anti-Plane Deformation . . . . . . . . . . 402.4.3 Plane Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.4.4 Axisymmetric Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.4.5 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

XII Contents

3 Some Elastic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.1 Longitudinal Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.1.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.1.2 Longitudinal Displacement of Strip . . . . . . . . . . . . . . . . . . 483.1.3 Deformation of Massif with Circular Hole

of Unit Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.1.4 Brittle Rupture of Body with Crack . . . . . . . . . . . . . . . . . 503.1.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.2 Plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.2.1 Wedge Under One-Sided Load . . . . . . . . . . . . . . . . . . . . . . 523.2.2 Wedge Pressed by Inclined Plates . . . . . . . . . . . . . . . . . . . 533.2.3 Wedge Under Concentrated Force in its Apex.

Some Generalisations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.2.4 Beams on Elastic Foundation . . . . . . . . . . . . . . . . . . . . . . . 613.2.5 Use of Complex Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 623.2.6 General Relations for a Semi-Plane Under

Vertical Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.2.7 Crack in Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.2.8 Critical Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.2.9 Stresses and Displacements Under Plane Punch . . . . . . . 683.2.10 General Relations for Transversal Shear . . . . . . . . . . . . . . 693.2.11 Rupture Due to Crack in Transversal Shear . . . . . . . . . . . 693.2.12 Constant Displacement at Transversal Shear . . . . . . . . . . 703.2.13 Inclined Crack in Tension . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.3 Axisymmetric Problem and its Generalization . . . . . . . . . . . . . . . 723.3.1 Sphere, Cylinder and Cone Under External

and Internal Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 723.3.2 Boussinesq’s Problem and its Generalization . . . . . . . . . . 743.3.3 Short Information on Bending of Thin Plates . . . . . . . . . 793.3.4 Circular Crack in Tension . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4 Elastic-Plastic and Ultimate State of PerfectPlastic Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.1 Anti-Plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.1.1 Ultimate State at Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . 854.1.2 Plastic Zones near Crack and Punch Ends . . . . . . . . . . . . 86

4.2 Plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.2.1 Elastic-Plastic Deformation and Failure of Slope . . . . . . . 884.2.2 Compression of Massif by Inclined Rigid Plates . . . . . . . 904.2.3 Penetration of Wedge and Load-Bearing Capacity

of Piles Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.2.4 Theory of Slip Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.2.5 Ultimate State of Some Plastic Bodies . . . . . . . . . . . . . . . 994.2.6 Ultimate State of Some Soil Structures . . . . . . . . . . . . . . . 1034.2.7 Pressure of Soils on Retaining Walls . . . . . . . . . . . . . . . . . 108

Contents XIII

4.2.8 Stability of Footings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.2.9 Elementary Tasks of Slope Stability . . . . . . . . . . . . . . . . . . 1114.2.10 Some Methods of Appreciation of Slopes Stability . . . . . 113

4.3 Axisymmetric Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1174.3.1 Elastic-Plastic and Ultimate States of Thick-Walled

Elements Under Internal and External Pressure . . . . . . . 1174.3.2 Compression of Cylinder by Rough Plates . . . . . . . . . . . . 1194.3.3 Flow of Material Within Cone . . . . . . . . . . . . . . . . . . . . . . 1214.3.4 Penetration of Rigid Cone and Load-Bearing Capacity

of Circular Pile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.4 Intermediary Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

5 Ultimate State of Structures at Small Non-Linear Strains . . 1255.1 Fracture Near Edges of Cracks and Punch

at Anti-Plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255.1.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255.1.2 Case of Crack Propagation . . . . . . . . . . . . . . . . . . . . . . . . . 1265.1.3 Plastic Zones near Punch Edges . . . . . . . . . . . . . . . . . . . . . 127

5.2 Plane Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1285.2.1 Generalization of Flamant’s Problem. . . . . . . . . . . . . . . . . 1285.2.2 Slope Under One-Sided Load . . . . . . . . . . . . . . . . . . . . . . . 1315.2.3 Wedge Pressed by Inclined Rigid Plates . . . . . . . . . . . . . . 1345.2.4 Penetration of Wedge and Load-Bearing Capacity

of Piles Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1395.2.5 Wedge Under Bending Moment in its Apex . . . . . . . . . . . 1405.2.6 Load-Bearing Capacity of Sliding Supports . . . . . . . . . . . 1445.2.7 Propagation of Cracks and Plastic Zones

near Punch Edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1465.3 Axisymmetric Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

5.3.1 Generalization of Boussinesq’s Solution . . . . . . . . . . . . . . . 1495.3.2 Flow of Material within Cone . . . . . . . . . . . . . . . . . . . . . . . 1515.3.3 Cone Penetration and Load-Bearing Capacity

of Circular Pile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545.3.4 Fracture of Thick-Walled Elements Due to Damage . . . . 156

6 Ultimate State of Structures at Finite Strains . . . . . . . . . . . . . 1616.1 Use of Hoff’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.1.1 Tension of Elements Under Hydrostatic Pressure . . . . . . 1616.1.2 Fracture Time of Axisymmetrically Stretched Plate . . . . 1626.1.3 Thick-Walled Elements Under Internal

and External Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1646.1.4 Final Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6.2 Mixed Fracture at Unsteady Creep . . . . . . . . . . . . . . . . . . . . . . . . 1666.2.1 Tension Under Hydrostatic Pressure . . . . . . . . . . . . . . . . . 166

XIV Contents

6.2.2 Axisymmetric Tension of Variable Thickness Platewith Hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

6.2.3 Thick-Walled Elements Under Internaland External Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

6.2.4 Deformation and Fracture of Thin-Walled ShellsUnder Internal Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

6.2.5 Thin-Walled Membranes Under Hydrostatic Pressure . . 1766.2.6 Two Other Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1786.2.7 Ultimate State of Anisotropic Plate

in Biaxial Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

Appendix B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Appendix C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Appendix D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

Appendix E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

Appendix F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

Appendix G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

Appendix H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

Appendix I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

Appendix J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

Appendix K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

Appendix L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

Appendix M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

Appendix N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

1

Introduction: Main Ideas

1.1 Role of Engineering Geological Investigations

An estimation of conditions of buildings and structures installation demandsa prediction of geological processes that can appear due to natural causes oras a result of human activity. This prediction should be based on a geolog-ical analysis which takes into account different forms of interaction betweencreated structures and an environment.

The prediction of the structures role is usually made by the methods of theengineering geology that includes computations according to laws of mechanics(deformation, stability etc.). The basic data are: geological schemes and cross-sections, physical-mechanical characteristics of soils, and others. An engineer-geologist participates in the choice of structures places and gives their proofs.

The engineer-geologist must always take into account that engineeringstructures made of concrete, brick, rock, steel, wood and other materialsshould not be considered separately from their foundations which must havethe same degree of reliability as a whole construction. The main causes oftheir destruction can be changes in their stress state and, consequently, – thedeformation of the soil. So, an engineer and a geologist must have enoughknowledge of the geo-mechanics as a part of the whole structures theory.

1.2 Scope and Aim of the Subject: Short Historyof Soil Mechanics

The Soil Mechanics deals mainly with sediments and their unconsolidated ac-cumulations of solid particles produced by mechanical and chemical disinte-gration of rocks. It includes the parts (according to another classification theyare the components of the geo-mechanics): 1. Mechanics of rocks. 2. That ofmantel or rigolitth – the Soil Mechanics itself. 3. Mechanics of organic masses.4. Mechanics of frozen earth.

2 1 Introduction: Main Ideas

The main features of soils are:

a) a disintegration which distinguishes a soil from a rock, the particles cannot be bonded but they form a body with the strength much lower thanthat of a particle; it induces a porosity that can change under externalactions;

b) they have the property of permeability;c) strength and stability of a soil is a function of a cohesiveness and of a

friction between the particles;d) the stress-strain dependence of a soil includes as a rule residual compo-

nents and influence of time (a creep phenomenon).

The main tasks of the Soil Mechanics are as follows. 1. The establish-ment of basic laws for soils as sediments and other accumulations of particles.2. The study of soil strength and stability including their pressure on retainingwalls. 3. Investigation of structures strength problems in different phasesof deformation.

For the solution of these tasks two main methods are used – theoretical (onthe basis of a mathematical approach) and modelling with different materials.Here the first of them is considered.

The development of the Soil Mechanics began at the end of the eighteenthcentury. The first period is characterized by a rare use of scientific methods.The theoretical investigations that are actual now were contained in the worksof French scientists (Prony, 1810; Coulomb, 1778; Belidor, 1729; Poncelet,1830 and others) who were solving the problem of soil pressure on a retainingwall, and Russian academician Fussa (the end of the eighteenth century) whoreceived a computational method for a beam on elastic foundation. Until thebeginning of the twentieths century works on the Soil Mechanics were linkedwith determination of a soil pressure on retaining walls and a solution of thesimplest problems of slopes and footings stability. V.I. Kurdumov began in1889 laboratory tests of soils as foundations of structures.

The next step in the Soil Mechanics was made by Carl Terzaghi /1/ in USAand N.M. Gersewanov in the USSR. They gave the schemes of deformation andultimate state calculations. In the problem of soil strength we must mentionthe works of N.N. Pouzyrevski in the USSR and O. Frolich /2/ in Germany.The books of N.A. Cytovich /3/, V.A. Florin, N.N. Maslov and other Russianscientists have broad applications. Rigorous solutions for a soil massif at itsultimate state was given by V.V. Sokooovski /4/.

1.3 Use of the Continuum Mechanics Methods

Computations in the Soil Mechanics are usually fulfilled by methods of theContinuum Mechanics. Although soils have different mechanical propertiestheir settlings have been found on the base of the Elasticity Theory /5/. Thecomplete solution of its plane problem was given by N.I. Muschelisvili /6/

1.3 Use of the Continuum Mechanics Methods 3

thanks to the use of the complex variables. An application of the PlasticityTheory methods are met much rarer although they give solutions nearer tothe reality /7/.

The founder of the strength disciplines is G. Galilei who in 1638 publishedhis book “Discorsi E Demonstrazioni Matematiche Intorno A Due NuoveScience” (Talks and proofs concerning two new sciences) in which he groundedthe Theoretical Mechanics and the Strength of Materials. In that time lawsof deformation were not discovered and G. Galilei appreciated a strength ofbodies directly.

The discovery of the linear dependence between acting force and inducedby it displacement by R. Hooke in 1676 gave the basis of the Elasticity Theory.Rigorous definitions of stresses and strains formulated a civil engineerO. Cauchy. The main mathematical apparatus of this science was intro-duced in the works of G. Lame and B. Clapeyron who worked at that periodin St.-Petersburg institute of Transport Communications. The number ofpractically important problems was solved thanks to the famous principleof Saint-Venant.

In calculations according to the scheme of the Elasticity Theory the maintask is the determination of stress and strain fields. An estimation of a strengthhas as a rule an auxiliary character since a destruction in one point or in agroup of them does not lead to a failure of a structure. The Galilei’s idea ofan appreciation of the strength of the whole body found applications only insome districts of the Continuum Mechanics (the stability of compressed barsaccording to the Euler’s approach, a failure of some objects in the StructuralMechanics, the theory of the ultimate state of soils, and quite recently – inthe theory of cracks propagation due to the Griffith’s idea /8/).

The computations according to the ultimate state began to develop thanksto a study of metal plastic deformation at the end of the nineteenth cen-tury by French investigators Levy, Tresca, Saint-Venant and at the beginningof the last century by German scientists Mises, Hencki, Prandtl. The latterintroduced the diagram of an ideal elastic-plastic material and solved a rowof important problems including geo-mechanical ones. The important role forthe practice play up to now two Gvozdev’s theorems of the ultimate stateof a plastic body /9/ (the static one that proposes the ultimate load as amaximum force among all corresponding to an equilibrium and a minimumone for all kinematically possible forms of destruction) which he proved inthe thirties of the last century in the USSR for the objects of the StructuralMechanics. In the Media Mechanics such theorems were formulated at thefifties in USA.

The prominent contribution to the Plasticity Theory made W. Prager,F. Hodge and A. Nadai who received also a row of important results in the geo-mechanics and other disciplines. The works of Soviet scientists V. Sokolovski,L. Kachanov and A.Il’ushin in this field are also well-known. The specialinterest have their investigations in the Theory of Plasticity of a hardeningmaterial which describes the real behaviour of a continuum and includes as


particular cases the linear elasticity and the ideal plasticity. An intensive studyof cracks in an elastic-plastic material provides G. Rice in USA.

The phenomenon of creep was discovered by physicists who (Boltzman,Maxwell, Kelvin) constructed in the nineteenth century the constitutive equa-tions which are actual now. In the technique the creep processes are studiedsince the twentieths of the last century in the connection with the metals de-formation at elevated temperatures under constant loads. The construction ofthe basic relations followed the ideas of the Plasticity Theory of a hardeningbody. The large work in this direction was made by F. Odquist.

In twentieths-thirtieths of the last century Odquist and Hencki found anopportunity to compute the fracture time of a bar in tension under constantload when its elongation tends to infinity. This idea began to spread out onlyafter the work of N. Hoff (1953) who used more simple equation of creep andreceived the good agreement with test data. To predict an earlier failure of thestructure L.M. Kachanov introduced in the sixtieths a parameter of a damageas a ratio of a destructed part of a cross-section to the whole one. Accordingto his idea the bar either elongates infinitely or is divided in parts when thedefects fit up the whole area.

Another way to describe the ultimate state in a creep opens the crite-rion of infinite elongations rate at the beginning of unstable deformation(R. Carlsson, 1966). The introduction in constitutive equations of a damageparameter allowed S. Elsoufiev to find on the base of the criterion the ultimatestate of many objects including geotechnical ones /7.10/.

1.4 Main Properties of Soils

1.4.1 Stresses in Soil

Due to the weight (which is always present), tectonic, hydrodynamic, physical-chemical, residual and other processes internal stresses appear in the earth.In a weightless massif at an action of load P (Fig. 1.1) in a point M a part ofthe body under cross-section nMl is in an equilibrium with internal stressesp which are distributed non-uniformly in the part of the massif. If they areconstant in a cross-section the relation for their determination (Fig. 1.2) is

p = P/A (1.1)

where A is an area of cross-section aa.The non-uniformly distributed stresses can be found according to

expressionp = lim

dA→0(dP/dA). (1.2)

Here dA is an elementary area in a surrounding of the investigated point anddP – the resultant of forces acting in it.

1.4 Main Properties of Soils 5

P

Mn l

dP

dA

Fig. 1.1. Stresses in massif of soil

P

A

a a

p

Fig. 1.2. Stresses in compressed bar

The value and the direction of stress p depend not only on a meaningof external forces and the position of the point but also on the direction ofa cross-section. If vector p is inclined to a plane it can be decomposed intonormal σ and shearing τ components (Fig. 1.3).

Since materials resist differently to their actions such a decomposition hasa physical meaning. In the general case an elementary cube is cut around thepoint on each side of which one normal and two shearing stresses act (seeChap. 2 further).


M

p

a

a

τ

σ

Fig. 1.3. Decomposition of stress p

The base of a structure is a part of the massif where stresses depend onthe structure erected. It differs from a foundation that transfers the structureweight to the base. The boundary of the latter is a surface where the stressesare negligible. All the artificial soil massifs (embankments, dams etc.) are notthe bases, they are structures.

Stresses in the massif under external loads differ from their real meaningson values of the soil self-weight components. These so-called natural pressuresdepend on a specific weight γe of the soil, a coordinate z of the point and thedepth of an underground water. The natural pressure is determined by relation

σ′z = γez + γ ′z′

where γ ′ is the specific weight of the soil with the consideration of its sus-pension by the underground waters, z′ is the depth of the point from theirmirror. Normal stresses on vertical planes are determined as

σ′x = σ′

y = ζσ′z.

Here ζ = ν/(1 − ν) is the factor of lateral soil expansion and ν – thePoisson’s ratio.

1.4.2 Settling of Soil

Phases of Soil State

An elastic solution shows that with a growth of a load an a punch plasticdeformation begins at its edges. Then inelastic zones expand according tothe plastic solution. Experiments confirm this picture if we suppose that the


P

cb

cb

z

a

Fig. 1.4. Zones of stress state under punch

S

A

h

rock

p

ho

Fig. 1.5. Settling of layer of limited thickness

part of the soil (districts a, b in Fig. 1.4) acts together with the punch. Atthe same time the expansion of the soil upwards (zones c in the figure) takesplace and slip lines appear in zones b. This phenomenon was observed byV.I. Kourdumov in his tests. Professor N.M. Gersewanov proposed to considerthree stages of the base state at the growth of the load: 1) its condensation,2) an appearance of the shearing displacements and 3) its expansion. By theseprocesses a condensed solid core (zone a in the figure) takes place and it movestogether with the punch making additional plastic districts.

Settling of Earth Layer of Limited Thickness

Under an action of uniformly distributed load p at a large length (Fig. 1.5)an earth layer is exposed to a pressure without lateral expansion. The


process is similar to the compressive deformation and the problem becomesone-dimensional. If the layer is supported by an incompressible and impen-etrable basis its full settling is equal to the difference of initial and currentlengths, that is

S = ho − h. (1.3)

The skeleton volume in a prism with a basic area A before and after thedeformation remains constant as

Aho/(1 + eo) = Ah/(1 + e) (1.4)

where eo, e are factors of the soil porosity before and after the loading. Theyare computed as ratios of pores and skeleton volumes.

Solving (1.4) relatively to height h and putting it into (1.3) we find

S = ho(eo − e)/(1 + eo). (1.5)

Now we introduce a factor a as

a = (eo − e)/p

and put it in (1.5) which gives

S = hoap/(1 + eo). (1.6)

Valuea(1 + ao)

is a factor av of soil compressibility and (1.6) becomes

S = hoavp. (1.7)

As a result we receive that the full settling of the soil layer under a homo-geneous loading and in the conditions of an absence of its lateral expansionis proportional to the thickness of the layer, the intensity of the load anddepends on the properties of the soil.

Role of Loading Area

As natural observations show a settling depends on the loading area in aform of the curve in Fig. 1.6. on which three districts can be distinguished:1 – of small areas (till 0.25m2) when the soil is in the phase of the shearingdisplacements and the settling decreases with a growth of A, a zone 2 where thesoil is in the phase of a condensation and the settling is practically proportionalto A0.5 as

S = Cp√

A


0

1

2

3S

0.5 5 7 10 A0.5

Fig. 1.6. Dependence of settling on area of footing

where C is a coefficient of proportionality, and part 3 in which with a fall ofthe condensation role of the core the divergences from the proportionality lawis observed. We must also notice that the relation above is valid at pressureswhich do not exceed the soils practical limit of proportionality and at itsenough homogeneity on a considerable depth.

At the same area of the footing, pressure and other equal conditions thesettling of compact (circular, square etc.) foundations is smaller than stretchedones. That follows also from analytical solutions (see further). At a transferfrom square footing to rectangular one (at equal specific pressure) the activedepth of the soil massif increases.

Influence of Load on Footing

At successive increase of a load on a soil three stages of its mechanical state areobserved – of condensation 1 (Fig. 1.7), shearing displacement 2 and fracture 3.In the first of them the earth’s volume decreases and deformation’s rate fallswith a tendency to zero. In this stage the dependence between the acting forceF and the settling can be described by the Hooke’s law:

S = Fh/EA. (1.8)

where E – modulus of elasticity.The second stage is characterized by an appearance of shearing displace-

ment zones with growth of which the settlings become higher and their ratedecreases more slowly. In the third stage strains increase rapidly and soilexpanses out of a footing. Deformation grows catastrophically and the set-tlings are big.

At cycling loading the soil’s deformation increases with the number ofcycles. Its elastic part changes negligibly and the full settling tends to aconstant value. In the last state the soil becomes almost elastic.


O F

1

2

3

S

Fig. 1.7. Influence of load on footing

O

S

t

1

2

Fig. 1.8. Settling in time

Influence of Time

The experiments with earth and natural observations show that at a constantload a development of the settling in time can be represented by Fig. 1.8. Curve1 corresponds to sands in which settling happens fast since the resistance tosqueezing a water out is small. Case 2 takes place in disperse soils such asclays, silts and others in which pores in natural conditions are filled with thewater. The rate of the soil’s stabilization depends on its water penetrationand a creep of the skeleton.

The settling does not end in a period of a structure’s construction andcontinues after it. The time at which the full deformation takes place dependson the consolidation of a layer under the footing. In its turn the last phe-nomenon is determined by a rate and a character of external loading andby properties of soil, firstly by its compressibility and ability to water pene-tration. In conditions of good filtration the settling goes fast but at a weakpenetration the process can continue years.


O t

1

2

3S

Fig. 1.9. Combined influence of time and loading

Combined Influence of Time and Loading

At small loads F the settling grows slowly (curve 1 in Fig. 1.9) and tends toa constant value. There is a maximum load on footing at which this processtakes place. At bigger F the settling increases faster (line 2 in the figure)at approximately constant velocity and it can lead to a failure of structures(curve 3 in Fig. 1.9).

The velocity of the deformation influences the strength of structures sincethey have different ability to redistribute the internal forces at non-homogeneous settlings of a footing. At high velocities of the settling thebrittle fractures can take place, at slow ones – creep strains. For soils in whichpores are fully filled with the water the theory of filtration consolidation isusually used.

1.4.3 Computation of Settling Changing in Time

Premises of Filtration Consolidation Theory

The initial hypothesis of the theory is an assessment that the velocity of thesettlings decrease depends on an ability of the water to penetrate the soil.Above that the following suppositions are introduced:

a) the pores of a soil are fully filled with water, which is incompressible,hydraulically continuous and free,

b) earth’s skeleton is linearly deformable and fully elastic,c) the soil has no structure and initially an external pressure acts only on

the water,d) a water filtration in the pores subdues to the law of Darcy,e) the compression of the skeleton and a transfer of the water are vertical.

Model of Terzaghi-Gersewanov

The model is a vessel (Fig. 1.10) filled with water and is closed by a suckerwith holes. It is supported by a spring which imitates a skeleton of the soil


p

2 1dz

z

h

Fig. 1.10. Terzaghi-Gersewanov’s model

and the holes – capillaries in earth. If we apply to the sucker an external loadthen its action presses in an initial moment only the water. After some timewhen a part of the water flows out of the vessel the spring begins to resist toa part of the pressure. The water is squeezed out slowly and has an internalpressure as

H = p/γw (1.9)

where γw is a specific gravity of the water.The filtration of the water subdues to the Darcy’s law, which is however

complicated by a presence of a connected part of it. At small values of thehydraulic gradient a filtration can not. overcome the resistance of the waterin the pores. Its movement is possible only at an initial value of the gradient.

Differential Equation of Consolidation due to Filtration

In the basis of the theory the suppositions are put that a change of the waterexpenditure subdues to the law of filtration and a change of a porosity isproportional to the change of the pressure.

Now we consider a process of the soil compression under a homogeneouslydistributed load. We suppose that in an initial state the soil’s massif is ina static state which means that a pressure in pores is equal to zero. Wedenote the last as pw (zone 2 in the figure) and effective pressure which actson the solid particles as pz (zone 1 in the figure). At any moment the sum ofthese pressures is equal to the external one as

p = pw + pz. (1.10)

In time the pressure in the water decreases and in the skeleton – increasesuntil the last one supports the whole load.

1.5 Description of Properties of Soils and Other Materials 13

At any moment an increase of the water expenditure q in an elementarylayer dz is equal to the decrease of porosity n that is

∂q/∂z = −∂n/∂t. (1.11)

The last expression gives the basis for the inference of differential equation ofa consolidation theory which with consideration of (1.9) is

∂H/∂t = Cv∂2H/∂z2 (1.12)

where Cv = K∞/avγw is a factor of the soil’s consolidation.Relation (1.12) is a well-known law of diffusion and it is usually solved in

Fourier series (we have used this approach particularly for the appreciation ofwater and air penetration through polymer membranes).

For the determination of a settling in a given time a notion of consolidationdegree is usually used. It is defined as the ratio between the settling in theconsidered moment and a full one or

u = Sf/S. (1.13)

It can be found through the ratio of areas of pressure diagrams in the skeleton(pz) at the present moment and at infinite time as

u =

h∫

0

(pz/Ap)dz (1.14)

where Ap is the area of fully stabilized diagram of condensed pressure.Putting in (1.13) expression for pressure pz in the soil’s skeleton which is

received at the solution of equation (1.12) we have after integration

u = 1 − 8(e−N + e−9N/9 + e−25N/25 + . . ...)/π2 (1.15)

where e – the Neper’s number, N = π2Cvt/4h2 – a dimensionless parameterof time. Putting (1.15) in (1.7) we find the settling at given time t.

For bounded, hard plastic and especially for firm consolidated soils,containing connected water the theory can not be used.

In the conclusion we must say that (1.15) is the creep equation which isdifficult to apply to boundary problems and for this task simpler rheologicallaws can be used. So, this way is considered in the book.

1.5 Description of Properties of Soils and OtherMaterials by Methods of Mechanics

1.5.1 General Considerations

Usually problems of the Soil Mechanics are divided in two main parts.The first of them deals with the settling of structures due to their own

weight and other external forces. This problem is almost always solved by the


methods of the Elasticity Theory although many earth massifs show non-linearresidual strains from the beginning of a loading.

The second task is the stability which is connected with an equilibrium ofan ideal soil immediately before an ultimate failure by plastic flow. The mostimportant problems of this category are the computation of the maximumpressure exerted by a massif of soil on elastic supports, the calculation ofthe ultimate resistance of a soil against external forces such as a verticalpressure acting on an earth by a loaded footing etc. The conditions of a lossof the stability can be fulfilled only if a movement of a structure takes placebut the moment of its beginning is difficult to predict.

So, we must consider the conditions of loading and of support required toestablish the process of transition from the initial state to a failure. Here wedemonstrate on simple examples the approaches of finding the ultimate statein a natural way that can help us to study this process.

1.5.2 The Use of the Elasticity Theory

Main Ideas

Some earth components and even their massifs (rock, compressed clays, frozensoils etc.) subdue to the Hooke’s law i.e. for them displacements are propor-tional to external forces almost up to their fracture without residual strains.Here in a simple tension (Fig. 1.11) stress p which is determined by relation(1.1) and is equal in this case to normal component σ is linked with relativeelongation

ε = l/lo − 1 (1.16)

where index refers to initial values (broken lines in the figure) by the lawsimilar to (1.8) as follows

σ = Eε (1.17)

Here according to the main idea of the book we show the use of thisexpression for the prediction of a failure of some structure elements.

F

A

l

p

Fig. 1.11. Bar in tension


qds

M

Q

M

Q*

x

ds

Fig. 1.12. Element of bar under internal and external forces

Some Solutions Connected with Stability of Bars

As was told before one of the first methods of theoretical prediction of failuregave L. Euler for a compressed bar. His approach can be generalized andwe give the final results. We begin with static equations of a part of it(Fig. 1.12) as

dδQ/ds + χoxδQ + δχxQo = 0,dδM/ds + χoxδM + δχxMo + ixδQ = 0 (1.18)

and constitutive equation similar to (1.17):

oM = BjEδχ (j = x, y, z). (1.19)

Here i, Q, M are vectors – unit, of shearing force, of bending moment,Q∗ = Q + dQ, M∗ = M + dM, δ – sign of an increment, χ – curvature of thebar, x – sign of multiplication, BjE – rigidity and subscript denote initialvalues (before a failure).

For a ring with radius r and thickness h under external pressure q we haveMo = χxo = χyo = Qyo = Qzo = 0, χyo = 1/r,Qxo = −qr,ds = rdθ where z,x are normal and tangential directions, θ is the second polar co-ordinate. Forthe planar loss of stability we have from (1.18), (1.19)

d2δQz/dθ2 + (1 + qr3/ByE)δQz = 0.

The solution of this problem is well-known and according to perioditycondition 1+qr3/ByE = n2 (n = 1, 2, 3, . . .). The minimum load takesplace at n = 2 which gives

q∗ = 3EIy/r3.

This result is valid for a long tube if we replace moment of inertia Iy relativelyto axis y by cylindrical rigidity h3/12(1−ν2) where h – thickness of the tube,ν is the Poisson’s ratio. The solution is used for the appreciation of the strengthof galleries in an earth.

For a bar under compression and torsion (Fig. 1.13) the solution of (1.18),(1.19) can be presented in form

(M/M∗)2 + F/F∗ = 1 (1.20)


F

M

l2r

MF

Fig. 1.13. Circular bar under compression and torsion

q

l l dl

r

q

max p

Fig. 1.14. Crack under tension

where values M∗ = 2πEI/l,F∗ = π2EI/l2 refer to the separate actions of therespective loads, I = πr4/4 is axial moment of inertia of a circle. The relationcan be used for the appreciation of strength of bores, piles, drills etc.

Bases of Crack Mechanics

The basis of the modern linear mechanics of cracks was given by Englishscientist A. Griffith /8/. In order to clear up the idea we consider a plate(Fig. 1.14) with a narrow crack of length 2l perpendicular to tensile stresses qin infinity. The distribution of the latter near the crack edge is shown in theleft part of the picture and their maximum can be given by expression

max p = 2q√

l/r (1.21)


where r is a radius of curvature in the end of the crack When the last onebegins to propagate the following amount of the work is freed:

dW = (max p)2(r/E)dl

or with consideration of (1.21)

dW = 4(q2/E)ldl. (1.22)

This process is resisted by the forces of surface stretching with an energy dU =2γsdl where γs is the energy per unit length. In the critical state dU = dWand we derive from (1.22) the value of a critical stress:

q∗ =√

γsE/2l (1.23)

The more detailed analysis of this theory will be given in Chaps. 3–5.

1.5.3 The Bases of Ultimate Plastic State Theory

Main Ideas

Many soils and other materials have small hardening and angle of internalfriction as well as negligible deviation from the condition of constant volume.For them the ultimate state of structures according to the scheme of an ideal(perfect) elastic-plastic body can be used. Its diagram in coordinates p, eis given in Fig. 1.15 with p∗ an yielding point. In the continuum mechanicsconsequent coordinates are σ, ε and σyi.

To catch the idea of the method we consider a part of a beam (Fig. 1.16, a)loaded by moments M with a cross-section on Fig. 1.16, b. Diagram of stress inelastic state with yielding value at most remote point of the compressed partof the cross-section is given in Fig. 1.16, c. With the growth of the momentσ-diagram in the compressed zone becomes a trapezoid. Then the yielding

p

A

O

B

e

p*

Fig. 1.15. Diagram of ideal elastic-plastic material


M M

D

p* p*

p*

y

x z

ya) b) c) d)

C

Fig. 1.16. Pure bending of beam

point is reached in the tensile zone and the σ-diagram there becomes also atrapezoid.

The distribution of stresses at ultimate state is drawn in Fig. 1.16, d wherewe have two rectangles. From equilibrium condition we receive expression

ΣX = σyiA+ − σyiA− = 0

which gives A+ = A− = A/2. It means that in ultimate state tensile andcompressed areas are equal to one half of the whole one. From the secondstatic equation (ΣMz = 0) we find the ultimate load as

M∗ = σyiA(CD)/2

where (CD) is the distance between centroids of compressed and tensile zones.Particularly for a rectangle with a width b and a height h we compute

M∗ = σyibh2/4. (1.24)

Here we must underline that we used for the final results only the horizontalpart of the diagram in Fig. 1.15 and so the method can be also applied to abrittle fracture.

Ultimate State of Statically Indeterminate Beams

As the second practical example of the theory we consider a retaining wallor a piles sheet under triangular load of a soil or a liquid (Fig. 1.17, a). Theapproximate M-diagram in elastic state is given in Fig. 1.17, b. In the ultimatestate plastic hinges with moments M∗ (see relation (1.24) at b = 1) appear.According to M-diagram one of them (see the broken line in Fig. 1.17, a) is inthe fixed end (point A), another – somewhere in the span (point C). In orderto find distance x we use static equations


H

x

lC

A

M* qa) b)

M*

Fig. 1.17. Statically indeterminate beam

−MA = ql2/6 − Hl = M∗,MC = Hx − qx3/6l = M∗.

Excluding from these two expressions reaction H we have

q = 6M∗/x(l − x).

According to the second Gvozdev’s theorem /9/ the position of hinge Cgives minimum to q-value. So, we compute

x = 1/2, q∗ = 24M∗/l2.

The same value of q∗ can be got according to the first Gvozdev’s theorem asa maximum load in the ultimate static state /7/.

Ultimate State of Plates in Bending

We consider firstly a polygonal plate (Fig. 1.18, a) with simply supportededges. We model its part ODCE as a double-supported beam AB (Fig. 1.18,b) with a M-diagram similar to the broken axis (broken line in Fig. 1.18, c)under concentrated load in point O which acts on the plate. The value ofMmax is evident

Mmax = MC = Fd(l − d)/l. (1.25)

Taking Mmax equal to M∗ according to relation (1.24) at b = 1, substitutingin (1.25) d = L tan ϕs, l − d = L tan ψs where ϕs = a,ψs = c in Fig. 1.18, a,s = 1, 2, . . .n (n is a number of plate’s corners), and summarising the momentsalong the rays of fracture from point O to the corners of the plate we find

F∗ = M∗n∑

s=1

(cot ϕs + cot ψs). (1.26)


C

C

F*

LE

D

A B

Eca

L

O

O

OB

B

A

A

D

d l-d

a)

b)

c)

Fig. 1.18. Polygonal plate in bending

For a rectangle with a width b, a height h we have cot ϕs = h/b, cot ψs =b/h,n = 4 and we compute from (1.26)

F∗ = 4M∗(h/b + b/h).

In the case of a square (h = b) it gives

F∗ = 8M∗. (1.27)

For a right polygon with n corners and the force F in its centre we haveϕs = ψs = π(0.5 − n−1), cot ϕs = cot ψs = tan(π/n) and according to (1.26)

F∗ = 2nM∗ tan(π/n).

When n → ∞ we derive for a circle

F∗ = 2πM∗. (1.28)

If the plate has fixed edges we can suppose that the fracture occurs alsoalong them. From Fig. 1.19 where part ODE of the plate is represented wehave the static equation ΣMDE = 0

F∗H = M∗l

and since d = Hcot ϕs, l−d = Hcot ψs we get again (1.26) and so for our case

F∗ = 2M∗n∑

s=1

(cot ϕs + cot ψs). (1.29)


d l− dD E

a c

H

O F*

Fig. 1.19. Equilibrium of part of plate

For a circle with the critical force in its centre we have similar to (1.28)

F∗ = 4πM∗ (1.30)

and it is less than in the case of a right polygon F∗ = 4nM∗ tan(π/n) and soits mode of fracture is a circle of an indeterminate radius.

If the distributed load q is applied to the plate we can suppose that itdoes the same work as its resultant F. So, in the case of q = constant we mustreplace in the previous relations force F∗ by q∗A/3 where A is an area of theplate’s surface. So, for a right polygon with supported and fixed ends as wellas for a circle we have respectively

q∗ = 6M∗/R2, q∗ = 12M∗/R2 (1.31)

where R is radius of an internal circumference. The first (1.31) is valid for asquare 2Rx2R.

The use of the first Gvozdev’s theorem (the static one) for plates will begiven later in Chap. 3.

1.5.4 Simplest Theories of Retaining Walls

We consider a development of main ideas of the ultimate state computation onthe example of a retaining wall loaded mainly by a soil. The first investigatorsof this problem supposed that in an ultimate state triangle ABC (Fig. 1.20, a)acts against the wall. After somewhat naıve work of Belidor (1729) Coulomb/11/ considered the earth pressure on the portion BC of the vertical side CEand assumed that the earth has a tendency to slide down along some plane AB.Neglecting any friction on CB he concluded that reactions of the wall withtheir resultant H are horizontal. Weight Q of prism ABC is (γeh

2 tan Ψ)/2where γe is a specific weight of the earth. Resultant R of the reactions along


C A

R

H

a) b)

R

H

Q Q

B

E

h

nj

Ψ + j

Ψ

Ψ

Fig. 1.20. Retaining wall

sliding plane AB forms friction angle ϕ with normal n to AB. It means thatR is inclined to a horizontal direction under angle Ψ + ϕ.

The triangle in. Fig. 1.20, b represents the condition of an equilibrium ofprism ABC from which we have

H = Q cot(Ψ+ϕ) = 0.5γeh2 tan Ψ cot(ϕ+Ψ) = 0.5h2γe(1−f tan Ψ)/(1+f cot Ψ)

(1.32)where f = tan ϕ is a coefficient of friction. Then he found the maximum of Hand equalling its derivative by Ψ to zero he received expression

tan2 Ψ = (1 − f tan Ψ)/(1 + f cot Ψ)

from which it followstan Ψ = −f +

√1 + f2. (1.33)

Prony threw equation (1.33) in simpler form, viz cot(Ψ+ϕ) = tan Ψ whichleads to

Ψ = 0.5(π/2 − ϕ). (1.34)

All above was concerned an active pressure of a soil on the retaining wall.Similar computations for a passive state of an earth when the wall movesagainst it gives relation like (1.34)

Ψ = 0.5(π/2 + ϕ). (1.35)

Rankine /12/ offered a method of finding proper dimensions of a retain-ing wall. He considered a horizontal plane (Fig. 1.21) when σx, σy are mainstresses. It allows to construct the Mohr’s circle (Fig. 1.22) in coordinates τ, σ.In ultimate state ϕ is an angle of repose, so from the figure we have

σy − σx = (σy + σx) sin ϕ


x

y

τσ

Ψ

σx

σy

Fig. 1.21. Stress state near retaining wall

2Ψt

j

s m

t

s = sqs1 = sx

s3 = sy

srs O

c

Fig. 1.22. Mohr’s circle

or

σy/σx = (1 + sin ϕ)/(1 − sin ϕ). (1.36)

But since σy = γey the maximum horizontal reaction on the retaining wall inultimate equilibrium state is

σx = γey(1 − sin ϕ)/(1 + sin ϕ). (1.37)

Rankine recommends this pressure as to be used when investigating a stabilityof a retaining wall. In the case of a passive pressure similar computations give

σx = γey(1 + sin ϕ)/(1 − sin ϕ). (1.38)

Relations (1.37), (1.38) are very often used in design practice (see Chap. 4for example). But it is not difficult to show that they coincide with theCoulomb’s law (1.32) if we put there the value of Ψ according to (1.34),(1.35). So, we can conclude that the Rankine’s angle of repose is equal to theCoulomb’s angle of friction and their theories are the same.


q

h

b

a

y

x

Fig. 1.23. Depth of retaining wall

An approach for determining the necessary depth of foundation was pro-posed by Pauker (St.-Petersburg, 1889) – Fig. 1.23. Considering an element abunder the wall and using equation (1.37) of the Rankine’s theory he concludedthat at the moment of sliding the lateral pressure σx must satisfy equation

σx = q(1 − sin ϕ)/(1 + sin ϕ). (1.39)

Taking into account relation (1.38) for an ultimate state of the soil with depthh instead of y and combining it with (1.39) he got the necessary depth of thefoundation according to expression

h = (q/γe)(1 − sin ϕ)2/(1 + sin ϕ)2. (1.40)

To verify this equation some interesting experiments were provided byV. Kourdumov in the laboratory of the Ways Communication institute inSt.-Petersburg and by H. Muller-Breslau in 1906. These tests showed thatthe pressure on the wall can sometimes be higher than that predicted by theCoulomb’s theory.

1.5.5 Long-Time Strength

All the materials deform in time and this phenomenon is called a creep. Thecreep curves for soils are shown in Figs. 1.8, 1.9 and in the continuum me-chanics they are considered in coordinates ε, t at different σ. We can con-ditionally distinguish three parts of creep- a primary one with d2ε/dt2 < 0,the second district where dε/dt is approximately constant and the third onewith d2ε/dt2 > 0. The last one is usually linked with fracture due to a damage(a development of internal defects) and a decrease of cross-section (at tension).


N. Hoff /13/ used for the second portion of the curves a power law

dε/dt = Bσm (1.41)

where B = constant and m is an exponent of the hardening law, to predictthe rupture time at tension which corresponds to infinite elongation of a barunder constant load F. He used also a link between conditional σo = F/Ao

and true σ = F/A stressesσ = σoeε (1.42)

which follows from the condition of constant material’s volume Ao1o = Al(Fig. 1.11). Here

ε = ln(l/lo). (1.43)

is a true strain which can be got as a sum of its increments related to currentlengths.

Putting (1.42) into (1.41) he received after integration in limits 0 ≤ t ≤ t∞,0 ≤ ε ≤ ∞ a finite value of rupture time as

t∞ = (B(σo)mm)−1. (1.44)

According to this relation a structure can be destroyed at any load. It makesconditional the traditionally used strength limits. The scheme can be gener-alized.

Experiments show that for creep curves the following law may be intro-duced /7, 10/

εe−αε = Ω(t)σm (1.45)

where Ω is an experimentally determined function of time and factor e−αε(α ≥0) takes into account a damage increase that induces a decrease of ultimatestrain and time for not enough ductile materials, a growth of their volume,the third parts of creep curves and other effects.

Tests show that relation (1.45) is valid for monotonous loading while strainε does not diminish. At stepwise or interrupted change of stress (Fig. 1.24)rheological law (1.45) can be used for the parts where σ is not less than onprevious ones if the time is calculated from a beginning of a new application ofstress. The agreement of (1.45) with an experiment is better for more unsteadycreep /14/. It can be applied to finding the ultimate state of a structure.

Putting (1.42) into (1.45) and using the criterion /15/ dε/dt → ∞ we geton critical strain and time as

ε∗ = (α + m)−1,Ω(t∗) = ((α + m)(σo)me)−1. (1.46)

If a change of the bar’s dimension is small or true stress σ is constant we havefrom (1.46) the fracture values due to damage only as

ε∗ = 1/α,Ω(t∗) = (ασme)−1. (1.47)

Similarly from (1.46) we can find at α = 0 critical values for perfectly duc-tile bodies. When an influence of time is negligible (Ω = constant) relation


e

s o

s Is III

sII = 0

s IV

O ttI tII tIII tIV

Fig. 1.24. Stepwise and interrupted loading

(1.45) becomes a constitutive equation of the Plasticity Theory developed hereand expressions (1.46), (1.47) generalize the approach of maximum loads atunstable deformation.

Expressions (1.45). . .(1.47) give an opportunity to find all the parametersin rheological law (1.45). When we take from the primary parts of creep curvesat the same time strains ε we can construct diagrams σ(ε) approximation ofwhich by the power function allows to find the exponent m of the hardeninglaw. Dividing ε by σm we have a creep curve Ω(t) which can be also approxi-mated by a power law e.g. Ω = Btn. Damage parameter α can be found in twoways, either according to the first expressions (1.46), (1.47) or by approxima-tion of the third parts of creep curves. Very often the both approaches givenear values of α.

1.5.6 Eccentric Compression and Determination of CreepParameters from Bending Tests

We take a bar with a width 2h (Fig. 1.25, a) and unity thickness loaded bycompressive force F with eccentricity e. We use the law of plane cross-sectionsε = ky1 (broken line in the figure) where k is a curvature and two systems ofcoordinates – xOy and xO1y1 with y1 = y + c. If we apply forces in point Owe find axial load N = F and bending moment M = Fe.

We take a material of non-linear creep type (1.45) at α = 0 as

p ≡ σ = ω(t)εµ (1.48)

where ω = Ω−µ, µ = 1/m (0 ≤ µ ≤ 1). At ω =constant and µ = 1 (thenω = E) we have the Hooke’s law (1.17) – straight line OA in Fig. 1.15. Ifµ = 0 we receive the perfect plastic body with ω = σyi = p∗ – horizontalstraight line in the same figure.

With consideration of the expression above for ε we derive from (1.48) –Fig. 1.25, b.


c

O O

M

+

−

N

x

e F

k

h y−h1

a)

p

b)

Fig. 1.25. Eccentric compression of bar

p = ωkµ(y1)µ = ωkµ(y + c)µ (1.49)

Putting (1.49) into static equations ΣX = 0,ΣMo1 = 0 we compute

N = ωkµ

h−c∫

h+c

(y1)µdy1 = ωkµ((h + c)1+µ − (h − c)1+µ)/(1 + µ), (1.50)

M = ωkµ

h+c∫

h−c

(y1)1+µdy1 − Nc

= ωkµ((h + c)2+µ + (h − c)2+µ)/(2 + µ) − Nc. (1.51)

Now we consider some particular cases for ideal materials that are well-knownfrom a literature on the Strength of Materials.

If µ = 1, ω = E we have from (1.50), (1.51) N = 2Ekch,M = 2Ekh3/3and from (1.49) – a well-known formula (generalization of this law for brittlematerial see in Appendix A)

p = N/2h + 3My/2h3.

When p at y = h reaches its yielding point p∗ we receive the condition (straightline in Fig. 1.26 in coordinates M/p∗h2, N/2p∗h)

N/2p∗h + 3M/2p∗h2 = 1. (1.52)

At µ = 0, ω = p∗ we receive from (1.50), (1.51) N = 2p∗c,M = p∗(h2−c2)and after the exclusion of c we have (the curve in the figure)

M/p∗h2 + (N/2p∗h)2 = 1. (1.53)


0

2/3

2*

xh

M/p

1 N / 2 P*h

Fig. 1.26. Ultimate state of bar in eccentric compression-tension

F1/3

F

x

F F F

1/3

v

I II

1/31/3

y

M

Fig. 1.27. Deflection of beam

It is interesting to notice a similarity of equations (1.53) and (1.20) that showsa resemblance of the behaviour of structures in an ultimate state.

If c = 0 we find from (1.50), (1.51) N = 0,M = ωkµI which gives ageneralization of approximate bending equation

v′′ = −Mm/(ωI)m (1.54)

where I = 2h2+µ/(2+µ) – moment of inertia relatively to axis z perpendicularto directions x, y in Fig. 1.27, v – deflection of a beam and sign ′ denotes thederivative by coordinate x.

Function v(x) can be found by the same procedure as in the Strength ofMaterials and the Structural Mechanics for linear material. However, here wecan not use a single equation for the whole beam and must consider its partsseparately. We shall demonstrate it on the scheme which is often used in tests


for a study of materials behaviour (Fig. 1.27). Since the beam is symmetricwe can consider only its two portions. So, at (F/ωI)m = K we have in parts Iand II

v′′I = −Kxm, v′

I = v′I(0) − Kxm+1/(m + 1), vI = v′

I(0)x − Kxm+2/

(m + 1)(m + 2),v′′II = −K(1/3)m, v′

II = v′II(0) − Kx(1/3)m, vII = vII(0) + v′

II(0)x

− Kx2(1/3)m/2.

(1.55)

Here we take vI(0) = 0. From condition v′II(1/2) = 0 we find

v′II(0) = Klm+1/3m2.

From border demand v′I(1/3) = v′

II(1/3) we compute

v′I(0) = Klm+1(1/2 − m/3(m + 1))/3m.

Now the similar condition vI(1/3) = vII(1/3) gives the last unknown as

vII(0) = −K(1/3)m+2m/2(m + 2).

So, all the constants in (1.55) are found and the displacement and the angleof rotation in any point of axis x are determined. Particularly, the deflectionin the middle of the beam where it is usually measured is

vII(1/2) = Klm+2(1/4 − m/9(m + 2))/3m2. (1.56)

At m = 1 we compute well-known result v(1/2) = 23Fl3/648EI. Approxima-tion of (1.56) gives m, ω and hence Ω(t).

2

Main Equations in Media Mechanics

2.1 Stresses in Body

The stresses p, σ, τ for tension and compression were introduced in Chap. 1.In order to study the stress state in a point the cube is usually cut about it(Fig. 2.1). Acting on its faces stresses are often written in a form of a matrixwhich is called the tensor (of stresses) as follows

Tσ =

⎛⎝σxx τxy τyz

τyx σyy τyz

τzy τzy σzz

⎞⎠ (2.1)

where τij = τji and as a rule it is supposed σxx ≡ σx, σyy ≡ σy, σzz ≡ σz.There is an interesting interpretation of a tensor as the vector in

9-dimensional space /7/.Tensor (2.1) fully defines the stress state around a point because unit force

pn at any cross-section through this point with normal n(nx, ny, nz) can befound from static equations (Fig. 2.2 where stresses from Fig. 2.1 act on theopposite sides of the cube) as

pnx = σxnx + τxyny + τxznz,

pny = τxynx + σyny + τyznz,

pnz = τxznx + τyzny + σznz.

(2.2)

The projection of pn on n gives normal stress σn as

σn = σx(nx)2 + σy(ny)2 + σz(nz)2 + 2τxynxny + 2τxznxnz + 2τyznynz.

Components τnm and τnl can be determined in the same manner. The trans-formation of all the tensor stresses can be written in the form (i, j = x, y, z;r, s = n, m, l)

σrs =∑i

∑j

σkl cos(i, r) cos(j, s). (2.3)

32 2 Main Equations in Media Mechanics

y

X

Z

syy

sxx

szz

tyz

tyx

txy

txz

tzx

tzy

Fig. 2.1. Cube with stresses an its faces

y

z

x

pnz

pny

pnx

pn

n

Fig. 2.2. Decomposition of vector pn

When the cube rotates in a body stresses on its planes change and sucha position of it can be found that shearing components become zero. Thecorresponding faces and normal stresses on them are called main planes andmain stresses respectively. They can be found from conditions

(nx)2 + (ny)2 + (nz)2 = 1, (2.4)∣∣∣∣∣∣σx − σn τxy τxz

τxy σy − σn τyz

τxz τyz σz − σn

∣∣∣∣∣∣ = 0. (2.5)

The last of them gives equation relatively to σn

(σn)3 − I1(σn)2 + I2σn − I3 = 0 (2.6)

where n = 1, 2, 3, σ1 ≥ σ2 ≥ σ3 – main stresses and I1, I2, I3 – invariants; theydo not depend on the position of the cube and are computed as

2.2 Displacements and Strains 33

I1 = σx + σy + σz = σ1 + σ2 + σ3, (2.7)

I2 = σxσy+σzσy+σxσz−(τxy)2−(τxz)2−(τyz)2 = σ1σ2+σ2σ3+σ1σ3, (2.8)

I3 = σxσyσz + 2τxyτxzτyz − σx(τyz)2 − σy(τxz)2 − σz(τxy)2 = σ1σ2σ3. (2.9)

In the Mechanics of Fracture the fundamental role plays the maximum shear-ing stress that can be defined as follows

τe = 0.5(σ1 − σ3). (2.10)

From equations (2.2) with a help of Gauss-Ostrogradski theorem /6/ thedifferential static equations can be found: in the form

∂σx/∂x + ∂τxy/∂y + ∂τxz/∂z + X = 0,

∂τxy/∂x + ∂σy/∂y + ∂τyz/∂z + Y = 0, (2.11)∂τxz/∂x + ∂τyz/∂y + ∂σz/∂z + Z = 0

where X, Y, Z are volume’s forces. As we can see from (2.11) six stresses arelinked only by three expressions and the problem is statically indeterminate.To solve it the strains must be put in consideration.

2.2 Displacements and Strains

External forces deform a body, so its points move in new positions by dis-placements u(ux,uy,uz) which determine strains as /5/

εx = ∂ux/∂x, εy = ∂uy/∂y, εz = ∂uz/∂z,γxy = ∂ux/∂y + ∂uy/∂x, γxz = ∂uz/∂x + ∂ux/∂z, γyz = ∂uz/∂y + ∂uy/∂z.

(2.12)Here is also valid γij = γji, εxx ≡ εx, εyy ≡ εy, εzz ≡ εz and strains form atensor similar to (2.1):

Tε =

∣∣∣∣∣∣εx γxy/2 γxz/2

γxy/2 εy γyz/2γxz/2 γyz/2 εz

∣∣∣∣∣∣ . (2.13)

The strains are linked by well-known compatibility equations

∂2εx/∂y2 + ∂2εy/∂x2 = ∂2γxy/∂x∂y,

2∂2εx/∂z∂y = ∂(∂γxy/∂z + ∂γxz/∂y − ∂γyz/∂x)/∂x (2.14)

and 4 other similar relations can be received by cyclic change of indexes.The generalisation for finite linear strains can be fulfilled with consideration

of a change of measured basis (see (1.43)) as

εi = ln(li/lio) (i = x, y, z) (2.15)

and there is an option for shear /7/.


It is not difficult to notice an analogy between tensors (2.1) and (2.13)where σi, τij must be replaced by εi, γij/2 and vice versa. In this mannerinvariants for strains can be found from (2.7) . . . (2.9) and for maximum shearwe have

γm = ε1 − ε3. (2.16)

Now for 15 unknowns we have 9 equations and in order to close the systemwe must add some other 6 equations. They are constitutive laws which arenot the same for different materials.

2.3 Rheological Equations

2.3.1 Generalised Hooke’s law

In a certain range of deformation the Hooke’s law can be used. It is usuallywritten in form

εx = (σx − ν(σy + σz))/E, γzy = τzy/G,

εy = (σy − ν(σx + σz))/E, γxz = τxz/G (2.17)εz = (σz − ν(σx + σy))/E, γxy = τxy/G.

where modulus of shear G is linked with the Young’s modulus E and thePoisson’s ratio ν by expression

G = E/2(1 + ν). (2.18)

In practice other forms of the Hooke’s law are often applied. In order to getsome of them we summarize three left relations (2.17) and find

σm = Eem/3(1 − 2ν). (2.19)

Hereem = εx + εy + εz (2.20)

– a relative change of materials volume and

σm = (σx + σy + σz)/3 (2.21)

is the mean stress. Quantities σm, em are invariants of respective tensors (seeabove (2.7) and the analogy between them).

Taking off em/3 from the left three expressions (2.17) we receive relations

ei = (1 + ν)Si/E (2.22)

where ei, Si (i = x, y, z or 1, 2, 3 for main strains and stresses) are componentsof consequent deviators (tensors with sums of diagonal elements equal to zero).

2.3 Rheological Equations 35

Considering (2.22), (2.18) and three right equations (2.17) we can representthe Hooke’s law as

Sij = 2Geij (i, j = x, y, z) (2.23)

which must be used together with (2.19). Here for i = j Sij ≡ τij, eij ≡ γij/2.There are other forms of the Hooke’s law in the literature.

In practice the effective values σe, εe linked with the second invariantsof consequent deviators (for which expression (2.8) is valid with respectivereplacements) are used. They are usually written as follows

σe =√

3/2√

SijSij, (2.24)

εe =√

2/3√

eijeij. (2.25)

With consideration of (2.24), (2.25) relations (2.23) can be represented in theform useful for generalizations as

eij = (3εe/2σe)Sij. (2.26)

2.3.2 Non-Linear Equations

In the theory of ideal plasticity expressions (2.26) mean only a proportionalityof the deviators since ratio εe/σe is taken constant. Above that condition

σe = σyi (τe = τyi = σyi/2), (2.27)

where σyi was given in Chapter 1 and τyi is an yielding point in shear, togetherwith demand of the constant material’s volume is used.

For a hardening plastic body a dependence of εe on σe must be known andequation (2.26) becomes

eij = (3εe(σe)/2σe)Sij. (2.28)

The power law for this purpose is often applied in form

εe = Ω(σe)m. (2.29)

Here Ω, m are constants which can be found from tests in tension, compressionor bending (see Chap. 1). Here below expression (2.29) will be also used as adependence of stresses on strains:

σe = ω(εe)µ (2.30)

where ω = Ω−µ(1 ≥ µ ≥ 0).Replacing in (2.28) strains eij by their rates we get a power law for a steady

creep (see also (1.41))


deij/dt = (3B/2)(σeq)m−1Sij. (2.31)

Here σeq is an equivalent stress that takes into account an influence of a stressstate type on curves in coordinates σe, εe.

The simplest way to describe an unsteady creep with damage is to gener-alise relation (1.45) (for incompressible body) as /7/

exp(−αεeq)εij = (3Ω(t)/2)(σeq)m−1Sij. (2.32)

where εeq – an equivalent strain determine a development of a damage. Theexperiments /14/ show that as εeq the biggest main strain ε1 may be taken.If the influence of time is small (2.32) becomes the plasticity law (2.28) gen-eralized here.

2.3.3 Constitutive Equations for Anisotropic Materials

Equations (2.32) can be generalized for an anisotropic body /7/. If main axesof stress state coincide with orthotropic directions we have

εi exp(−αεeq) = Ω(t)(σeq)m−1(kj(σi − σk) + kk(σi − σj)). (2.33)

Here ks(s = i, j, k or x, y, z) are anisotropy factors. Their role can be appre-ciated by a divergence of curves in coordinates σe, εe (scalar properties of amaterial) and deviations from the condition µε = µσ (its vector properties)where µε, µσ are the Lode’s parameters (they are linked with the third invari-ants of respective deviators) which are determined by well-known relations/22/

µσ = (2σ2 − σ1 − σ3)/(σ1 − σ3) (1 ≥ µσ ≥ −1), (2.34)µε = 3ε2/(ε1 − ε3) (1 ≥ µε ≥ −1) (2.35)

where the latter expression is given for an incompressible body.Soils are usually stratified or laminated in horizontal directions. For this

case we can use a model of transversally isotropic body with vertical symmetryaxis z (Fig. 2.3). Putting into (2.33) ki = kj = 0.5 we have after transforma-tions

exp(−αεeq)εi = Ω1(t)(σeq)m−1(σi − kσj − (1 − k)σk),

exp(−αεeq)εk = Ω1(t)(σeq)m−1(2σk − σi − σj)(1 − k).(2.36)

where i, j = x, y; 2kk = k/(1 − k),Ω1 = Ω/2(1 − k) and the value of k can befound from the results of tension or compression in i-direction at σj = σk = 0according to relation k = −εj/εi.

When k = 0.5 the material is isotropic. If k = 1 it can be modelled as asystem of fibres parallel to axis k in a feeble matrix (a wood, for example).Cases k = 0 and k = −1 can be interpreted as systems of unconnected bars


z

x

y

k = 0

k =1

k = −1

Fig. 2.3. Cube with axis of symmetry z

in directions of axes i, j or under angles π/4 to them respectively (Fig. 2.3).The last two models correspond to a play-wood.

In the case of plane deformation we have from (2.36) at εy = 0 σy =kσx + (1 − k)σz and hence

exp(−αεeq)εz = − exp(−αεeq)εx = Ω1(t)(σeq)m−1(σz − σx)(1 − k2). (2.37)

Similar expressions can be found from (2.33) at the same supposition εy = 0and if kx = ky = 0.5. We can see that law (2.37) differs from (2.32) by aconstant multiplier.

Now we study the rheological properties of a transversally isotropic mater-ial in a plane stress state for three options of interposition of loading andisotropy planes. Using relations (2.24), (2.25) and (2.36) we receive

exp(−αεeq)εe = 2Ω1(3(1 − n + n2))−1/2√

Λ(σeq)m−1σe (2.38)

where n = σy/σx and the structure of function Λ(k,n) depends on a mutualposition of symmetry and loading planes.

For the case of Fig. 2.3 at σk = 0 we have from the first expression (2.36)

exp(−αεeq)εi = Ω1(t)(σeq)m−1(σi − kσj)(i, j,= x, y). (2.39)

In this option

Λ = (1 − k + k2)(1 + n2) + n(k2 − 4k + 1). (2.40)

Curves σe(εe) according to (2.38), (2.40) are represented in Fig. 2.4 by solid(k = 0.5), broken (k = 1), interrupted by points (k = 0) and dotted (k = −1)lines for n equal to 0 and 1. It is easy to see that the body k = 1 has anabsolute rigidity at n = 1 and small resistance to deformation at n = 0. This


n = 1

n = 1

n = 1n = 0

n = 0

0.5

k = 1

k = −1

k = 0

σe

εe

Fig. 2.4. Curves σe(εe) at different n and k when z is axis of symmetry

result is well-known from compression tests of timber along and transversethe fibres respectively. We must also notice that when parameter k diminishesthe material’s rigidity falls.

According to (2.34), (2.35) we find for the cases ε2 = εy and ε2 = εz

respectively

µε = 3(µσ ± (1 − 2k))/(5 − 4k ± (1 − 2k)µσ), (2.41)µε = 3(1 − k)(3 ± µσ)/(1 + k)(µσ ± (−1)) (2.42)

at upper signs. The corresponding curves are shown by solid and broken linesin Fig. 2.5. From this picture we can see that the influence of k on the vectorproperties of the body is also high and condition µε = µσ fulfils for an isotropicbody and in relation (2.42) at µε = µσ = −1.

A similar situation takes place in two other options of isotropy and loadingplanes interposition when we derive from (2.36)

exp(−αεeq)εi = Ω(t)(σeq)m−1(1 − k)(2σi − σj),

exp(−αεeq)εj = Ω(t)(σeq)m−1(σj − (1 − k)σi).(2.43)

In (2.43) parameter k may be determined according to relation k = −εz/εj atσi = 0 where j is axis of symmetry.

If i = x we have for Λ

Λ = 3(1 − k)2(1 − n) + n2(1 − k + k2) (2.44)

and function σe(εe) can be constructed as in Fig. 2.4 with the mutual replace-ment of curves n = 0 and n = 1. Expressions (2.42), (2.43) for ε2 = εy andε2 = εx can be used for lower signs and consequent diagram µε(µσ) can befound from Fig. 2.5 by rotating it about the centre at angle π.


−1

1

1

1

1

0

µ ε

−µσ

−1

−1

0.5

0.5

0.80.8

Fig. 2.5. Dependence µε(µσ) at different k when z is axis of symmetry

0

n = 0.5 n = 0.5n = 0;1

n = 0.5

n = 0;1

0.5σe

εeZ*

Fig. 2.6. Curves σe(εe) at different n and k when y is axis of symmetry

Such similarities of these two cases can be explained by the same positionof the plane with τe relatively to the symmetry axis.

If in (2.43) i = y is the axis of symmetry function Λ(n, k) can be found byreplacement in (2.38), (2.44) n by 1/n. Corresponding curves σe(εe) are givenin Fig. 2.6 by the same lines as in Fig. 2.4. In order to find all expressionsµε(µσ) we must consider three possibilities of relations between strains whenwe have

µε = 3µσ(1 − k)/(1 + k), (2.45)

µε = 3(±(−(1 + k)) − µσ(1 − k))/(1 + k ± µσ(k − 1)) (2.46)

for the options ε2 = εy, ε2 = εz (upper signs) and ε2 = εx (lower signs) respec-tively. The corresponding diagrams are given by solid,broken and interruptedby points lines in Fig. 2.7 and we can see also the high influence of k here.Option (2.45) embraces the first and the third quadrants of the plane when kchanges from −1 till 1 including straight line k = 0.5.


−1

−1

−1

−1 −1

1 1

µε

µσ

1

0

0.5

0

0.5

Fig. 2.7. Function µε(µσ) at different k when y is axis of symmetry

Following the idea of (2.33) we can generalize (2.31) for a steady creep ina similar way as

dεi/dt = B(σeq)m−1(kj(σi − σk) + kk(σi − σj)). (2.47)

and the option of transversally isotropic material can be also developed.

2.4 Solution Methods of Mechanical Problems

2.4.1 Common Considerations

In the common case of the continuum mechanics it is necessary to find 15functions from 15 equations ((2.11), (2.12) and of (2.17) type). Searched vari-ables should satisfy boundary conditions when stresses or strains are given onparts of body’s surface. The consequent problems are called the first and thesecond border tasks. A mixed problem appears when stresses and strains aregiven simultaneously on different parts of the surface.

The usual way to solve the problem consists in exclusion of variables,so only stresses or strains (displacements) remain. But up to date no finalresults are known for the general task even at ideal materials in considera-tion. For this reason simplifying hypotheses of geometrical character are in-troduced when the task is studied as the anti-plane (a longitudinal shear),plane, axisymmetric etc. problem.

2.4.2 Basic Equations for Anti-Plane Deformation

In this case we have only five unknowns (Fig. 2.8): τxz ≡ τx, τyz ≡ τy, γxz ≡γx, γyz ≡ γy and uz.

2.4 Solution Methods of Mechanical Problems 41

y

z

x

τy

τy τx

τx

Fig. 2.8. Stresses at anti-plane deformation

According to (2.11) . . . (2.13) they are linked by equations

∂τx/∂x + ∂τy/∂y = 0, (2.48)γx = ∂uz/∂x, γy = ∂uz/∂y, (2.49)

∂γx/∂y − ∂γy/∂x = 0. (2.50)

In elastic range we receive from (2.17)

τx = Gγx, τy = Gγy (2.51)

and for 5 unknowns we have 5 equations (2.48), (2.49) and (2.51) or for 4variables τx, τy, γx, γy − 4 equations: (2.48), (2.50) and (2.51).

For an ideal plastic body condition τe = τyi where

τe =√

(τx)2 + (τy)2 (2.52)

makes the problem statically determinate.Lastly at unsteady creep we have from (2.32)

exp(−αεeq)γi = Ω(t)(τe)m−1τi (2.53)

where i = x, y and maximum shear may be computed through γx, γy byexpression similar to (2.52):

γm =√

(γx)2 + (yγ)2. (2.54)

For some problems it is more convenient to use polar coordinatesr =

√x2 + y2, tan−1 θ = y/x (Fig. 2.9) where

τr ≡ τrz = τx cos θ + τy sin θ, τθ ≡ τθz = −τx sin θ + τy cos θ (2.55)


q

x x

yhr

O

Fig. 2.9. Polar coordinate system

and instead of (2.48) we have

∂(rτr)/∂r + ∂τθ/∂θ = 0. (2.56)

The shearing strains in this case are:

γr ≡ γrz = ∂uz/∂r, γθ ≡ γθz = ∂uz/r∂θ (2.57)

and compatibility equation (2.50) becomes

∂γr/∂θ − ∂(rγθ)/∂r = 0. (2.58)

Expressions (2.51) . . . (2.54) are valid at the replacement in them x, y by r, θ.

2.4.3 Plane Problem

Here 8 main variables do not depend on one coordinate (for example z) andfor them we have from (2.11), (2.12) and (2.14) common equations

∂σx/∂x + ∂τxy/∂y + X = 0, ∂τxy/∂x + ∂σy/∂y + Y = 0, (2.59)

εx = ∂ux/∂x, εy = ∂uy/∂y, γxy = ∂ux/∂y + ∂uy/∂x, (2.60)

∂2εx/∂y2 + ∂2εy/∂x2 = ∂2γxy/∂x∂y. (2.61)

The Hooke’s law (2.17) in the case σz = 0 (plane stress state) becomes

Eεx = σx − νσy,Eεy = σy − νσx,Gγxy = τxy. (2.62)

If εz = 0 (plane deformation) then we find from the third expression (2.17)σz = ν(σx + σy) and relations (2.62) are valid at the replacement in them


E, ν by E/(1 − ν2), ν/(1 − ν) respectively and it is not difficult to prove thatG-value remains the same.

Excluding from (2.59), (2.60), (2.62) stresses and strains we get 2 equa-tions for ux, uy. If we solve the problem in stresses we can receive from(2.59) . . . (2.62) a byharmonic equation for a function Φ as

∂4Φ/∂x4 + 2∂4Φ/∂x2∂y2 + ∂4Φ/∂y4 = 0 (2.63)

where

σx = ∂2Φ/∂y2, σy = ∂2Φ/∂x2, τxy = −∂2Φ/∂x∂y − yX − xY. (2.64)

The maximum shearing stress in this problem is

τe = 0.5√

(σx − σy)2 + 4(τxy)2 (2.65)

and here we can also conclude that for ideal plasticity condition τe = τyi =σyi/2 together with equations (2.59) makes the problem a statically determi-nate one. For hardening at creep body we have for the plane deformation

exp(−αεeq)εx = − exp(−αεeq)εy = 3Ω(t)(σeq)m−1(σx − σy)/4,

exp(−αεeq)γxy = 3Ω(σeq)m−1τxy. (2.66)

The main equations in the polar coordinate system (Fig. 2.9) can be de-rived directly or by transformations of corresponding expressions in Decart’svariables x, y. In /7/ a simplified procedure is used for this purpose (afterdifferentiation we equal θ to zero). As a result we receive firstly static laws as

r∂σr/∂r + σr − σθ + ∂τrθ/∂θ = 0, ∂(r2τrθ)/∂r + r∂σθ/∂θ = 0. (2.67)

Combining (2.67) we obtain a very useful expression

∂2τrθ/∂θ2 + ∂(r∂(σr − σθ)/∂θ)/∂r − r∂(r2τrθ)/r∂r)/∂r (2.68)

Strains are linked with displacements by relations similar to (2.60) andthey can be received from them with the help of vector transformationequations similar to (2.55):

εr = ∂ur/∂r, εθ = ur/r + ∂uθ/r∂θ, γrθ = r∂(uθ/r)/∂r + ∂ur/r∂θ. (2.69)

Condition of constant volume εr + εθ = 0 can be written with the help of twofirst expressions (2.69) as

∂(rur)/∂r = −∂uθ/∂θ (2.70)

and instead of (2.61) we have

r∂(∂(r2εθ)/r∂r)/∂r − ∂2εθ/∂θ2 = ∂2(rγrθ)/∂r∂θ. (2.71)


It is interesting to notice the identity of (2.71) with (2.68) if we replace inthe first of them εθ, γrθ by τrθ and σr − σθ respectively.

The Hooke’s law (2.62), relations (2.65), (2.66) are valid here with thereplacement in them x, y by r, θ. Transformation equations (2.4) can beexpressed here as (see also Fig. 2.9)

τrθ = 0.5(σy − σx) sin 2θ + τxy cos 2θ,

σr = σm + 0.5(σx − σy) cos 2θ + τxy sin 2θ,

σθ = σm − 0.5(σx − σy) cos 2θ − τxy sin 2θ

(2.72)

where (see the Mohr’s circle in Fig. 1.22)

σm = 0.5(σx + σy) = 0.5(σθ + σr). (2.73)

Lastly instead of (2.63), (2.64) we write

(∂/r∂r + ∂2/r2∂θ2 + ∂2/∂r2)(∂Φ/r∂r + ∂2Φ/r2∂θ2 + ∂2Φ/∂r2) = 0 (2.74)

and

σr = ∂Φ/r∂r + ∂2Φ/r2∂θ2, σθ = ∂Φ/∂r2, τ = −∂(∂Φ/r∂θ)∂r. (2.75)

2.4.4 Axisymmetric Problem

In this case cylindrical coordinates (Fig. 2.10) can be used and if all thevariables do not depend on angle θ the main equations are:

εr = ∂ur/∂r, εθ = ur/r, εz = ∂uz/∂z, γrz = ∂ur/∂z + ∂uz/∂r,

∂σr/∂r + ∂τrz/∂z + (σr − σθ)/r = 0, ∂τrz/∂r + ∂σz/∂z + τrz/r = 0,

∆σr − 2(σr − σθ)/r2 + 3(∂2σr/∂r2)/(1 + ν),

∆σθ + 2(σr − σθ)/r2 + 3(∂σr/∂r)/(1 + ν).

(2.76)

c

rsq

sc

sr

szsr

q

N

z

x

O

Fig. 2.10. Cylindrical and spherical coordinates


The first line in (2.76) is the Saint-Venant’s expressions, the second onerepresents static law and the latter two are compatibility equations in which∆ is the Laplace’s operator of this task for the solution in an elastic range

∆ = ∂2/∂r2 + r−1∂/∂r + ∂2/∂z2.

2.4.5 Spherical Coordinates

ρ∂σρ/∂ρ + ∂σχ/∂χ + 2σρ − σχ − σθ + τρχ cot χ = 0,

∂(ρ3τρχ)/ρ2∂ρ + ∂σχ/∂χ + (σθ − σχ) cot χ = 0.(2.77)

Combining (2.77) at σθ = σχ we receive a very useful equation

∂2(ρ2(σρ − σχ))/ρ∂ρ∂χ + ∂2τρχ/∂χ2 + ∂(τρχ cot χ)/∂χ − ρ(∂(∂ρ3τρχ)/ρ2∂ρ)∂ρ = 0.

(2.78)

Strains are linked with displacements by expressions

ερ = ∂uρ/∂ρ, εχ = ∂uχ/ρ∂χ + uρ/ρ, εθ = uρ/ρ + uχ(cot χ)/ρ,γρχ = ∂uρ/ρ∂χ + ρ∂(uχ/ρ)/∂ρ.

(2.79)

The rheological laws will be given later for a concrete task.If the main variables do not depend on angle χ the equations are simpler

and can be written as follows

d(ρ2σρ)/ρdρ − 2σθ = 0, ερ = duρ/dρ, εθ = εχ = uρ/ρ. (2.80)

The strains are linked by the conditions of compatibility and of a constantvolume:

dεθ/dρ + (εθ − ερ)/ρ = 0,d(ρ2uρ)/dρ = 0. (2.81)

When uθ = uχ = 0 the second and the third equations are valid. Relationfor γρχ as well as the compatibility law are

γρχ = duρ/ρdχ,dεχ/dχ = γρχ. (2.82)

3

Some Elastic Solutions

3.1 Longitudinal Shear


If a force Q acts along axis z (Fig. 3.1) we have from (2.56) a simple equation

d(rτr)/dr = 0 (3.1)

with an obvious solution τr = C/r where C = −Q/2π from the equilibriumequation ΣZ = 0 of the part of the cylinder in Fig. 3.1. Using the Hooke’s law(2.51) and the first expression (2.58) we can write the final result as follows

τr = −Q/2πr,uz = −(Q/2πG) ln r + uo (3.2)

where uo is a constant.Now we study the problem in complex variables z = x + iy, ζ = ξ + iη

etc. The convenience of such an approach follows from the fact that mainunknowns (displacements, stresses and, hence, strains) can be determined bycomplex variables and their derivatives. According to definition differentiationof a function

w(z) = ϕ(x, y) + iψ(x, y) (3.3)

is possible when its real (Re) ϕ and imaginary (Im) ψ parts satisfy well-knownequations of Cauchy-Riemann

∂ϕ/∂x = ∂ψ/∂y, ∂ϕ/∂y = −∂ψ/∂x (3.4)

If the function w(z) is known the main variables of the task can be foundby the following relations (we can prove the second of them using expressions(3.4), (2.48), (2.50), (2.51))

uz = ϕ(x, y)/G = (Re(w(z))/G, τx − iτy = (τr − iτθ)e−iθ = w′(z). (3.5)

48 3 Some Elastic Solutions

z

r

τr

Q

Fig. 3.1. Anti-plane deformation of cylinder

Q

I

-I

z

u o

Fig. 3.2. Displacement of strip

The latter equation is derived with the help of relation for vector compo-nents transformation (2.55) and sign ‘ means a derivative by z. We can checkexpressions above on the example of Fig. 3.1 for which the solution can begiven in the following way:

w(z) = −(Q/2π)lnz + Guo (3.6)

The convenience of the complex variables usage consists in the opportunityof conformal transformations application when solutions for simple figures(a semi-plane or a circle) can be transformed to compound sections /16/.

3.1.2 Longitudinal Displacement of Strip

To derive the solution of this problem by the conformal transformation ofresult (3.6) to the straight line −l, l (Fig. 3.2) with the help of the Zhoukovski’srelation which was deduced for an ellipse and here is used for our case as

ζ = 0.5(z + z−1),z

z−1= (ζ ±

√ζ2 − l2)/l (3.7)

3.1 Longitudinal Shear 49

we put the second of these expressions into (3.6) and using (3.5) we receive

w(ζ) = −Qln(ζ/l +√

(ζ/l)2 − 1)/π + Guo, τξ − iτη = −Q/π√

ζ2 − l2 = w′(z).(3.8)

Along axis ξ we compute as follows

At/ξ/ < l : uz = uo, τξ = 0, τη = Q/π√

l2 − ξ2,

At/ξ/ > l : uz = uo − (Q/Gπ) ln(ξ/l +√

(ξ/l)2 − 1), τη = 0, τξ = Q/π√

ξ2 − l2.(3.9)

In the same manner the displacement and stresses in any point of the massifcan be found. The most dangerous points are η = 0, /ξ/ = l and in order toinvestigate the fracture process there it is convenient to use decompositionζ − l = reiθ which gives according to expressions (3.5), (3.8) and (2.52)

uz = uo − (Q/Gπ)√

2r/l cos(θ/2),τrτθ

= Q/π√

2rl×− cos(θ/2)sin(θ/2) , τe = Q/π

√2rl.

(3.10)From the third relation (3.10) we can see that the condition τe = constantgives a circumference with a centre at /ξ/ = l where a fracture or plasticstrains should begin.

3.1.3 Deformation of Massif with Circular Hole of Unit Radius

In this case (Fig. 3.3) the boundary conditions are:

τη|ζ=∞ = τζ, τρ|ρ=1 = 0. (3.11)

We seek the solution in a form

τz

τz

ξ

h

r

Z

Fig. 3.3. Massif with circular hole


w′(ξ) =∞∑

n=0

An/ξn (3.12)

and the first condition (3.11) gives immediately A−1 = −iτζ. Now withthe help of (3.5) we rewrite the second (3.11) as Re(eiθw′(eiθ)) = 0 from thatwe have A1 = −iτζ and all other factors are equal to zero. So, we receive

w′(ζ) = −iτζ(1 + ζ−2),w(ζ) = −iτζ(ζ − ζ−1). (3.13)

For example at η = 0 we compute by (3.5)

τξ = uz = 0, τη = τζ(1 − ξ−2)

and τη(1) = 2τζ – the dangerous point.

3.1.4 Brittle Rupture of Body with Crack

This problem is very significant in the Mechanics of Fracture. In the liter-ature it is usually named as the third task of cracks. Its solution can bereceived by conformal transformation of the first relation (3.13) with the helpof Zhoukovski’s expression (3.7) in which the variables ζ and z are inter-changed. So, with consideration of condition τo = 2τζ/l we have (Fig. 3.4)

w′(z) = −τoz/√

l2 − z2,w(z) = ±τo

√l2 − z2. (3.14)

It is not difficult to prove that the first expression (3.14) can be got fromthe second (3.8) after replacing in it ζ, l,Q/π by 1/z, 1/l, τol respectively.

At x = 0 we determine from (3.14)

uz = ±(τo/G)√

l2 + y2, τy = τoy/√

l2 + y2, τx = 0. (3.15)

Along the other axis (x) we find similarly

τo

τo

θ

z

yII r

x

Fig. 3.4. Crack at anti-plane deformation

3.1 Longitudinal Shear 51

uz = ±(τo/G)√

l2 − x2, τy = 0, τx = −τox/√

l2 − x2(/x/<l),

uz = 0, τx = 0, τy = τox/√

x2 − l2(/x/>l).(3.16)

According to the Clapeyron’s theorem we can compute the work whichis done by stress τy at its decrease from τo to zero which corresponds to aformation of the crack as

W = τo

1∫

−1

uzdx = π(τo)2l2/2G. (3.17)

When the crack begins to propagate an increment of the work becomes equalto a stretching energy 4γsdl where γs is this energy per unit length. From thiscondition we find critical stress

τ∗ = 2√

γsG/πl. (3.18)

From the strength point of view stresses and strains in the edge of the crackare of the greatest interest. To find them we use the asymptotic approach as inSect. 3.1.2 that in polar coordinates r, θ (see Fig. 3.4) according to expressions(3.5), (3.14) gives

uz = τo

√2rl sin(θ/2), τr =

√l/2rτo sin(θ/2),

τθ =√

l/2rτo cos(θ/2), τe = τo

√l/2r.

(3.19)

From the fourth of these relations we can see that in this case condition τe =constant represents also a circumference with the centre in the top of crack.

Since the largest part of the energy concentrates near the crack edges wecan use expressions (3.19) for the computation of τ∗. When the crack growswe should put in the first relation (3.19) θ = π, r = dl − x and in the thirdone – θ = 0, r = x, then the increment of the work at the crack propagationis equal to that in (3.17) as follows

dW =

dl∫

0

τθ(0)uz(π)dx = ((τo)2/G)ldl

1∫

0

√(1 − ξ)/ξdξ

or after integrationdW = (τo

2/2G)πldl.Now we introduce an intensity factor K3 = τo

√πl and equalling dW to

the stretching energy 2γsdl we find K3∗ = 2(γsG)0.5 after that the strengthcondition may be written in form

K3 ≤ K3∗ (3.20)

where factor K3∗ is determined by the properties of the continuum and itsvalue can be found experimentally in elastic or plastic range /17/. The testsshow that the condition K3∗ = constant fulfils well enough for brittle bodiesonly. However equation (3.20) characterizes a resistance of the material to thecrack propagation (the so-called fracture toughness).


3.1.5 Conclusion

Problems of anti-plane deformation are ones of the simplest in the Mechanicsof Continuum and Fracture. But their solutions have practical and theoreticalvalue. Many processes in the earth (a loss of structures stability, landslidesetc.) occur due to shear stresses. Later on we will consider the problems abovein a non-linear range and the analogy between the punch movement and crackpropagation will be used for finding the solution of one of them when a resultof the other is known. Moreover a similarity between these results and onesin the plane deformation will be also of great importance.

3.2 Plane Deformation

3.2.1 Wedge Under One-Sided Load

In this case we suppose that stresses and strains do not depend on coordinate r(Fig. 3.5) and expressions (2.67), (2.69) become

dτ/dθ + σr − σθ = 0,dσθ/dθ + 2τ = 0,dεr/dθ = γ − C/G (3.21)

where C is a constant. Combining these relations with the Hooke’s law (2.62)(in polar coordinates) we receive equation

d2τ/dθ2 + 4τ = 4C

which has the solution with consideration of boundary condition τ(±λ) = 0in form

τ = Co(cos 2θ − cos 2λ) (3.22)

and from (3.21)σrσr

= C1 + Co(2θ cos 2λ ± sin 2θ). (3.23)

Finally boundary conditions σθ(−λ) = 0, σθ(λ) = −p give the values ofconstants as

Co = 0.5p(sin 2λ − 2λ cos 2λ)−1,C1 = −p/2 (3.24)

and according to (2.65)

τe = 0.5p√

1 − 2 cos 2θ cos 2λ + cos2 2λ/(sin 2λ − 2λ cos 2λ). (3.25)

The analysis of (3.25) shows that at λ = π/4 the maximum shearing stressis the same in the whole wedge and it is equal to p/2. At other λ>π/4 thisvalue is reached only on axis θ = 0 (interrupted by points line in Fig. 3.5).

In order to compute displacements we firstly determine the strains accord-ing to the Hooke’s law (2.62) as

εrεθ

= (−0.5p(1 − 2ν) + Co(2(1 − 2ν)θ cos 2θ ± sin 2θ)/2G,

γ = Co(cos 2θ − cos 2λ)/G.(3.26)

3.2 Plane Deformation 53

Dp

C

q = ϑ

q = λ

q = −λ

q = −ϑ

q

−q

B

r

A

0

Fig. 3.5. Wedge under one-sided load

a r

r/a

/q

0.5

2ql

1σ q

θ

λ

I

Fig. 3.6. Wedge pressed by inclined plates

Using (2.69), neglecting the constant displacement and excluding infinitevalues at r = 0 we receive

ur = r(−0.5p(1 − 2ν) + Co(2(1 − 2ν)θ cos 2λ + sin 2θ))/2G, (3.27)

uθ = r(pθ(1 − 2ν) + Co(cos 2θ − 2((1 − 2ν)θ2 + 2(1 − ν) ln r) cos 2λ)/2G.

For incompressible material (ν = 0.5) relations (3.36), (3.27) become muchsimpler.

3.2.2 Wedge Pressed by Inclined Plates

Common Case

Let plates move parallel to their initial position (broken straight lines inFig. 3.6) with displacement V(λ) = Vo. Then according to (2.70), (2.71) atuθ = V(θ) and (2.69) we receive

ur = U(θ) − V′, εr = −εθ = −U/r2, γ = U′/r2 − f(θ)/r (3.28)


where f = V + V′′ and from (2.66) for m = 1,Ω = 1/3G we have at τ ≡ τrθ

τ = G(U′/r2 − f/r), σr − σθ = −4GU/r2 (3.29)

that gives together with (2.67)

σθ = F(r) + (G/r)∫

fdθ, σr = F(r) + (G/r)∫

fdθ − 4GU/r2. (3.30)

Putting stresses according to (3.29), (3.30) into the first static law (2.67)we receive an equality

r3dF/dr − Gr(f ′ +∫

fdθ) = −G(U′′ + 4U)

both parts of which must be equal to the same constant, e.g. n andf ′ +

∫fdθ = 0. With the consideration of symmetry condition we determine

f = −Csin θ,U = −D cos 2θ − n/4G

where C, D are constants. In order to find n we use a stick demand /18/U(λ) = 0 which gives

U = −D(cos 2θ − cos 2λ),

and, consequently, - the stresses as

τ = G((2D/r2) sin 2θ + (C/r) sin θ),

σθ = A + 2(GD/r2) cos 2λ + (GC/r) cos θ, (3.31)

σr = A − 2(GD/r2)(cos 2λ − 2 cos 2θ) + (G/C/r) cos θ

where constants A, C, D should be determined from condition σθ(a, λ) = 0and integral static equations

λ∫

−λ

σr(a, θ) cos θdθ = 0,

a+1∫

a

σθ(r, λ)dr = −ql.

(3.32)

Putting in (3.32) stresses according to (3.31) we derive at θ = λ

σθ = −(q/Bo)(Λ(1 − a2/r2) cos 2λ + (1 − a/r) cos λ (3.33)

where

Bo = Λ(1 + a/l)−1 cos 2λ + (1 − (a/l) ln(l/a + 1)) cos λ (3.34)

andΛ = 3(cos λ − λ/ sin λ)/16 sin2 λ. (3.35)


r/a

−1

σθ /q

Fig. 3.7. Model of Retaining Wall

r/a

σq /q

10.5

Fig. 3.8. Model of two foundations

The diagrams σθ(r)/a) at l/a = 9 are given by solid lines in Figs. 3.6. . .3.8 forλ = π/6, λ = π/4 (a model of a retaining wall) and λ = π/2 (a flow of thematerial between two foundations) respectively.

From (3.31), (2.65) with consideration of D-value we compute maximumshearing stress as

τe = (q/Bo)(Λ2(cos 2λ − cos 2θ)2 + (Λ sin 2θ + sin θ)2)0.5. (3.36)

To find maximum of τe we use condition dτe/dθ = 0 which gives θ = 0 andequation

cos2 θ + (1/6)(4Λ cos 2λ + 1/Λ) cos θ − 1/3 = 0. (3.37)

Investigations show that at λ < π/3 an impossible condition cos θ > 1 takesplace and hence τe should be found from (3.36) at θ = 0 as follows

/τe/ = qΛ(cos 2λ − 1)/Bo.

But at λ > π/3 max τe is determined by (3.36) with θ from (3.37). Diagrammax τe(λ) is given in Fig. 3.9 by solid line.


0

1

2

0.5

3

max τe /q

π/6 π/3 λ

Fig. 3.9. Diagram max τe(λ)

Some Particular Cases

Besides this common solution it is interesting to study two simpleroptions: C = 0 (a compulsory flow of the material between immovableplates) and D = 0 (when plates move and the compulsory flow is negligible).In the both cases we use expressions for stresses (3.31), condition σθ(a, λ) = 0and the second static equation (3.32). For the first case we have

σθ = −q(1 − a2/r2)/B1, (3.38)

τe = (q/B1)(1 − 2 cos 2θ/ cos 2λ + 1/ cos2 2λ)0.5 (3.39)

whereB1 = 1 − 1/(1 + l/a).

Diagrams σθ(r/a) also at l/a = 9 are shown in Figs. (3.6), (3.7) by brokenlines. Condition dτe/dθ = 0 give demand sin 2θ = 0 with consequent solutionsθ = 0 and θ = π/2 but calculations show that only the first of them gives toτe the maximum value which is

max τe = (q/B1)(1 − 1/ cos 2λ)

Diagram max τe(λ) according to this relation is drawn in Fig. 3.9 by brokenline and we can see that at λ = π/4 it tends to infinity. For the case D = 0we derive in a similar way

σθ = −(q/B2)(1 − (a cos θ)/(r cos λ)), (3.40)

τe = (q/B2)(sin θ/ cos λ) (3.41)

whereB2 = 1 − (a/l) ln(1 + l/a)


The highest value of σθ is at θ = λ and diagrams σθ(r/a) also for l/a = 9 aregiven by pointed lines in Figs. (3.6). . . (3.8). Maximum τe takes place at θ = λand the consequent diagram is shown in Fig. 3.9 by interrupted by pointscurve which tends to infinity at λ = π/2. So we can conclude that diagramsσθ(r/a) in the options above are near to each other and are distributed evenlynear the value equal to unity but max τe – values can be high and plasticdeformations are expected in some zones.

Case of Parallel Plates

As an interesting particular case we consider a version of parallel plates(Fig. 3.10). We take uy as a function of y only and according to incompressibil-ity equation εx + εy = 0 as well as symmetry and stick (at y = h) conditionswe find

ux = 3Vox(h2 − y2)/2h3,uy = Voy(y2 − 3h2)/2h3 (3.42)

where Vo is a velocity of plates movement. Then we compute strain by relation(2.60) and stresses by the Hooke’s law (2.62) and static equations as follows

τxy = −3VoGxy/h3,

σx = 3GVo(3(h2 − y2) + x2 − l2)/2h3, (3.43)

σy = 3GVo(y2 − h2 + x2 − l2)/2h3.

Here condition σy(l,h) = 0 is used. Integral equilibrium law

1∫

−1

σy(x,h)dx = −P (3.44)

givesP = 2GVol3/h3. (3.45)

P

P

yh

h

x

II

Fig. 3.10. Compression of layer by parallel plates


3.2.3 Wedge Under Concentrated Force in its Apex.Some Generalisations

General Case

In this problem we have boundary conditions as σθ = τrθ = 0 at θ = ±λ.Since angle λ is arbitrary we can suppose that σθ and τrθ are absent at anyθ. That allows to seek function Φ from (2.75) in form

Φ = rf(θ) + C (3.46)

where C is a constant. Now we put (3.46) into (2.74) and with considerationof symmetry as well as static condition at αo = 0 (Fig. 3.11) we receive

σr = −P(cos θ)/r(λ + 0.5 sin 2λ). (3.47)

The strains and displacements can be determined as usual.The particular case λ = π/2 is of great interest when

σr = −2P(cos θ)/πr,

εr = −2P(cos θ)/Eπr, (3.48)εθ = 2Pν(cos θ)/Eπr.

Using relations (2.69) we find displacements ur, uθ /5/ which we write for theedge of the semi-plane as follows

ux ≡ ur = −P(1 − ν)r/2Eπ,uy ≡ uθ = C + (2P/πE)lnx (3.49)

where C is a constant.

Pαo

x

ry

θ

λλ

σr

Fig. 3.11. Wedge under concentrated force


η dη

y

l x

p γ eh

−l

ϑ

ϑ1

ϑ2

Fig. 3.12. Deformation under flexible load

Case of Distributed Load

If the load is distributed in interval (a, b) we replace in the relations above Pby pdη and integrate as follows

uy = uo + (2/πE)

b∫

a

p(η)ln(x − η)dη (3.50)

and for the case p = constant at −l ≤ η ≤ l (Fig. 3.12) we compute

uy = uo +(p/πE)((x/l+1) ln(x/l+1)+(1−x/l) ln(x/l−1)) (/l/ ≤ x). (3.51)

In the same manner stresses under a flexible load in an interval (−l, l) canbe found. We begin with the first relation (3.48) and according to (2.72) wewrite

σy = −(2P/πr) cos3 θ = −2Py3/πr4,τxy = −2Pxy2/πr4, σx = −2Px2y/πr4.

Now as before we replace P by pdη and summarize the loads as follows

σy = −(2p/π)

1∫

−1

(y2 + (x − η)2)−2dη

or after integration

σy = p(υ1 − υ2 + 0.5(sin 2υ1 − sin 2υ2))/π. (3.52)

Similarly we find

σx = p(υ1−υ2+0.5(sin 2υ2−sin 2υ1))/π, τxy = p(sin2 υ1−sin2 υ2)/π. (3.53)


According to (2.65) and (2.72) we compute maximum shearing stress andmain components as

τe = p sin(υ1 − υ2)/π,

σ1σ3

= p(υ2 − υ1 ± sin(υ2 − υ1))/π.(3.54)

The biggest τe is p/π and it realizes on the curve x2 + y2 = l2 (broken linein Fig. 3.12). Hence if strength condition τe = τyi is used a sliding along thiscircumference should be considered. This result was particularly applied toan appreciation of the earth resistance to mountains movement.

The First Ultimate Load

For the foundation with width 2l and depth h we must add load γeh outsidethe main one (p in Fig. 3.12). We suppose the hydrostatic distribution of thesoil’s weight σ3s = σ1s = γe(h + y), We take also compressive stresses aspositive and rewrite the second relation (3.54) in the following way

σ1σ3

= (p − γeh)(υ ± sin υ)/π + γe(h + y) (3.55)

where υ = υ1 − υ2. Now we put these stresses in the ultimate equilibriumcondition (see broken line in Fig. 1.22) in form

σ1 − σ3 = 2(σm + c cot ϕ) sin ϕ. (3.56)

That gives expression

((p − γeh)/π) sin υ − (((p − γsh)/π)υ + γe(h + y)) sin ϕ = c cos ϕ

from which we can find equation of the boundary curve where the first plasticstrains can appear

y = ((p − γsh)/πγe)(sin υ/ sin ϕ − υ) − c/γe tan ϕ − h. (3.57)

Now we use the condition dy/dυ = 0 which gives cos υ = sin ϕ or

υ = π/2 − ϕ (3.58)

Putting (3.58) into (3.57) we derive

ymax = ((p − γeh)/πγe)(cot ϕ + ϕ − π/2) − c(cot ϕ)/γe − h. (3.59)

If we decide to find a load at which plastic deformation does not begin in anypoint we must suppose ymax = 0 that leads to minimum load

minpyi = π(γeh + ccotϕ)/(cot ϕ + ϕ − π/2) + γeh

which is always called in the literature as the first critical load that was foundby professor Pousyrevski in 1929.


y

x

r ( x )

p ( x )

Fig. 3.13. Beam on elastic foundation

3.2.4 Beams on Elastic Foundation

We model such a foundation as a set of closely spaced separate springs(Fig. 3.13) which exert reactions r(x) = cv(x) where c is a modulus of anelastic bed and v – a deflection of a beam. If distributed load is p(x) totalforces acting on the beam are q = p − r and according to the Strength ofMaterial law q = −M′′ where M is linked with the deflection by expression(1.54) at m = 1 as

v′′ = −M/EI. (3.60)

Combining all these relations we receive linear differential equation of thefourth order for the beam on elastic foundation as follows

vIV + 4β4v = p(x)/EI (3.61)

where4β4 = c/EI

is a proportionality factor.The general solution of (3.61) is

v = Aeβx cos βx + Beβx sin βx + Ce−βx cos βx + De−βx sin βx + vp. (3.62)

Here A, B, C, D – constants and vp is a particular solution.Now we consider the important example of concentrated load P in the

origin of the coordinate system x, y (Fig. 3.14, a). In this case p(x) = 0 andhence vp = 0. Since displacement in infinity has finite value we must supposeA = B = 0. Constants C, D can be found from conditions v′(0) = 0 andv′′(0) = −Q/EI = −P/2EI where Q is a shearing force (see Sect. 1.5.2). As aresult we have the following solution

v = Pe−βx(cos βx + sin βx)/8β3EI (3.63)

(broken line in Fig. 3.14, a) with maximum deflection at x = 0 as

vmax = P/8β3EI (3.64)

and v = 0 in point x = 3π/4β.


dη

p(η)P

y

P8EIb 3

P/4bπ /4b

3 π /4b

a

x

b

a)

b)

η

Fig. 3.14. Concentrated load in origin

Using law (3.60) we find from relation (3.63) expression for bendingmoment (Fig. 3.14, b)

M = Pe−βx(cos βx − sin βx)/4β (3.65)

with maximum value at x = 0 as Mmax = P/4β and M = 0 in points x = π/4β,x = 3π/4β etc. (Fig. 3.14, b).

In order to appreciate the role of the elastic foundation we compare thisresult to a similar bending solution for a beam with supports at x = ±3π/4β(the moments there are equal to −20.5Pe−3π/4/4β = −0.007Pl and they can beneglected). According to well-known relations of the Strength of Materials forsuch a beam without elastic foundation Mmax = Pl/4 and vmax = Pl3/48EI.For our task β = 3π/2l and we conclude that the elastic foundation decreasesthe maximum moment in 5.7 and the maximum deflection in 17.4 times. So therole of the bed is significant.

Similarly to the previous paragraph we can consider a distributed load in(3.63), (3.65) respectively as

v(x) = (1/8β3EI)b∫a

e−β(x−η)p(η)(cos β(x − η) + sin β(x − η))dη,

M(x) = (1/4β)b∫a

p(η)e−β(x−η)(cos β(x − η) − sin β(x − η))dη(3.66)

at a ≤ η ≤ b.

3.2.5 Use of Complex Variables

General Expressions

Main equations of the plane problem are given by relations (2.59). . .(2.62).Similarly to the case of a longitudinal shear it is convenient to use herecomplex variables when law (2.63) becomes


∂4Φ/∂z2∂z2 = 0 (3.67)

where the line over a value means that we must change in it i by –i. Thesolution of (3.67) is obvious, containing two functions, e.g. F(z) and χ(z).When they are known stresses and displacements can be found according torelations

σx + σy = 4ReF′(z),

σy − σx + 2iτxy = 2(zF′′(z) + χ′(z)),(3.68)

2G(ux + iuy) = κF(z) − zF′(z) − χ(z). (3.69)

Here κ is equal to 3−4ν and (3−ν)/(1+ν) for plane deformation and the samestress state respectively.

Functions F(z) and χ(z) must be found from border demands that for thefirst boundary problem (see sub-chapter 2.4) is

σy + iτxy = F′(z) + zF′′(z) + F′(z) + χ′(z) (3.70)

or as a resultant of forces on an arc ab:

Px + iPy = −(F(z) + zF′(z) + χ(z))|ba . (3.71)

Tension of Plate with Circular Hole

As an example we study a plane with a circular hole of a radius R (Fig. 3.15)under tension by stresses σ in infinity (the problem was solved by G. Kirschin 1898). According to border condition (3.71) we have at z = Reiθ

F(z) + zF′(z) + χ(z) = 0. (3.72)

R

x

yr θ

σ

σ

σ y

Fig. 3.15. Tension of plate with circular hole


Similar to the case of massif with circular hole at anti-plane deformation weseek the solution in series as

F(z) = Az + Fo(z), χ(z) = Bz + χo(z)

where A, B – constants and

Fo(z) =∞∑

n=0

An/zn, χo(z) =∞∑

n=0

Bn/zn. (3.73)

From the first expression (3.68) we find at z → ∞A = σ/4,B = σ/2. Thenwe put Fo(z), χo(z) into (3.72) that gives

σz/4 + Fo(z) + σz/4 + zF′o(z) + σz/2 + χo(z) = 0. (3.74)

Now we consider in (3.74) condition z = R2/z and compute

Fo(z) + R2F′o(z)/z + χo(z) = −σR2/2z − σR2/2z. (3.75)

Taking into account A, B – values above and integrating the derivative F′o(z) =

σR2/2z2 we finally receive

F(z) = σz/4 − σR2/2z, χ(z) = −σR2/2z + σz/2 − σR4/2z3. (3.76)

From (3.69) we have

2G(ux + iuy) = (κ − 1)σz/4 − κσR2/2z − σzR2/2z2 − σzR2/2z2 − σz/2

+ σR2/2z + σR4/2z3. (3.77)

To determine the stresses we differentiate functions F′(z), χ(z) as

F′′(z) = −σR2/z3, χ′(z) = σ/2 + σR2/2z2 + 3σR4/2z4. (3.78)

Then relations (3.68). . .(3.70) can be used to determine stresses and displace-ments. For example at y = 0 we have

σy = σ(1 + R2/2x2 + 3R4/2x4) (3.79)

(shaded diagram in Fig. 3.15) with maximum σy at x = R as

maxσy = 3σ. (3.80)

If x = 0, y = R we find in a similar manner τxy = 0, σx = −σ and hence it isthe dangerous point at uni-axial compression of a wall or a tunnel.

3.2.6 General Relations for a Semi-Plane Under Vertical Load

As in this case τxy = 0 at y = 0 we have from (3.70)

zF′′(z) = −χ′(z), χ(z) = F(z) − zF′(z). (3.81)


So, instead of the second expression (3.68) and (3.69), (3.70) we receive

σy − σx + 2iτxy = −4iyF′′(z),

σy + iτxy = F′(z) + F′(z) − 2iyF′′(z),(3.82)

2G(ux + iuy) = κF(z) − F(z) − 2iyF′(z). (3.83)

From (3.83) and the second (3.82) we have at y = 0

uy = (κ + 1)ImF(xo)/G, σy = 2ReF′(xo)

and it is not difficult to notice that they differ only by a constant multiplierfrom border conditions

uz = Rew(xo)/G, τz = −Imw′(xo)

(see relations (3.5)) for an anti-plane deformation. That allows to receivesolutions of plane problems by replacing in similar results of longitudinalshear functions w′(z) by −2iF′(z).

3.2.7 Crack in Tension

Replacing in the first expression (3.14) τo by –iσ/2 we get the solution fora plane with crack of length 2l perpendicular to tensile stresses σ in infinity(Fig. 3.16) in form

F′(z) = σz/2√

z2 − l2,F(z) = σ√

z2 − l2/2. (3.84)

If z → ∞ then F′(z) = σ/2, F′′(z) = 0 and from (3.83), (3.84) we haveσx = σy = σ instead of σx= 0, σy = σ. So, the solution is wrong. But for thepurpose of the further theory it is not important because stress σx does notinfluence the moment of a crack propagation beginning. Now since

F′′ = −0.5σl2(z2 − l2)−3/2 (3.85)

y

x

θ

σ

σ

r

Fig. 3.16. Crack in tension


y

P

x

P

Fig. 3.17. Crack under concentrated forces

we find according to (3.82), (3.83) at y = 0, x ≤ /l/

σy = 0,uy = (σ/4G)(k + 1)√

l2 − x2. (3.86)

In a similar way the case in Fig. 3.17 can be considered where wehave /17/

F′(z) = Pl/2πz√

z2 − l2,F(z) = (P/2π) cos−1 l/z

and it is an example of a stable crack because its length increases with agrowth of the forces P.

3.2.8 Critical Strength

Now we find the energy due to the creation of a crack. According to theClapeyron’s theorem (3.17) and relation (3.86) we compute

W = (σ/2)

l∫

−l

uydx = (1 + κ)πl2σ2/8G

and increment of the energy due to a propagation of the crack

dW = (1 + κ)πl(σ∗)2dl/4G (3.87)

must be equal to stretching resistance of the material 4γsdl. That gives thecritical value of σ

σ∗ = 4√

Gγs/πl(1 + κ) (3.88)

which differs from (1.23) by a constant multiplier depending on the type ofthe problem (plane deformation or the same stress state) and the Poisson’sratio.

As in Sect. 3.1.4 we can find σ∗ according to the relations of the asymptoticapproach z − l = reiθ which gives with a help of transformation equations like(2.55), (2.72)


ur + iuθ = e−iθ(ux + iuy), σθ − σr + 2iτrθ = (σy − σx + 2iτxy)e2iθ (3.89)

as well as (3.82). . .(3.84) and (2.65)

σθ = (K1/√

2πr) cos3(θ/2), σr = (K1/√

2πr)(1 + sin2(θ/2)) cos(θ/2),

τrθ = (K1/2√

2πr) sin θ cos(θ/2), τe = (K1/2√

2πr) sin θ,

(3.90)

ur = (K1/G)√

r/2π(0.5(κ + 1) − cos2(θ/2)) cos(θ/2),

uθ = (K1/G)√

r/2π(0.5(1 − κ) − sin2(θ/2)) sin(θ/2)(3.91)

whereK1 = σ

√πl (3.92)

is stress intensity coefficient for the first crack task. Constructing the equilib-rium equation in its usual form

dl∫

0

uθ(dl − r, π)σθ(r, 0)dr = 2γsdl

and using (3.90), (3.91) we find

K1∗ = 4√

Gγs/(κ + 1) (3.93)

and the strength condition is

K1 ≤ K1∗.

There are some other approaches to a strength computation for brittle media.G. Barenblatt (see /17/) introduced a crack in a form of a beck (Fig. 3.18). Hismodel gives strength value in 1.27 times higher than the Griffith’s relation.Similar to that idea was introduced by N. Leonov and V. Panaciuk /19/. Intheir approach the crack begins to propagate when opening δ reaches its criti-cal value δcr (Fig. 3.18). In this moment the critical stress can be computed as

σ∗ =√

Eσbrδcr/πl (3.94)

where σbr is a limit of brittle strength.

dcr

sbr

Fig. 3.18. Model of G. Barenblatt


P

x

yτe = const

uy

σy

Fig. 3.19. Pressure of punch

3.2.9 Stresses and Displacements Under Plane Punch

M. Sadowski solved this problem (Fig. 3.19) using the analogy method /20/.Replacing in the second relation (3.8) Q, ζ, w′(ζ) by P, z, 2iF′(z) respectivelywe receive

F′(z) = −P/2π√

l2 − z2,F(z) = −(Pi/2π) ln(z +√

z2 − l2) + 2Guo. (3.95)

We can easily notice that this result can be got from the first expres-sion (3.84) after the consequent replacement of z, l, σ, by 1/z, 1/l, −Pi/πlrespectively.

With a help of (3.82), (3.83) we find a distribution of stresses (broken linefor σy in Fig. 3.19) and displacement uy (solid curves outside the punch) aty = 0 at x < l, x > l respectively as

uy = uo, τxy = 0, σx = σy = −P/π√

l2 − x2, (3.96)

σx = σy = τxy = 0,uy = uo − (P/2πG)(1 + κ) ln(x/l −√

(x/l)2 − 1). (3.97)

The computations show that diagram uy outside the punch is near to thatone for uniformly distributed load according to (3.51).

In a similar way as before we find with a help of (3.95), (3.82), (3.83) and(2.65) in the asymptotic approach

σrσθ

= −(P/π√

2rl)(1 ± cos2(θ/2)) sin(θ/2),

τrθ = (P/2π√

2rl) sin θ sin(θ/2), τe = (P/2π√

2rl) sin θ,

(3.98)

uθ = u1 − (P/πG)√

r/2l(0.5(κ + 1) − sin2(θ/2)) cos(θ/2),

ur = u2 − (P/πG)√

r/2l(0.5(κ − 1) + cos2(θ/2)) sin(θ/2)(3.99)

where u1,u2 – constants. It is easy to notice that τe in this task differs fromthat in the problem of crack (see the fourth relation (3.90)) by a constant


multiplier. The condition τe = constant is shown by pointed line in the leftpart of Fig. 3.19 under the edge of the punch and the plastic zone must havethis form.

3.2.10 General Relations for Transversal Shear

In this case we have on axis x condition σy = 0 and from (3.68) we find aftersome simple transformations

χ′(z) = −(2F′(z) + zF′′(z)), χ(z) = −(F(z) + zF′(z)). (3.100)

Putting these expressions into (3.68), (3.69) we receive

σy − σx + 2iτxy = −4(F′(z) + iyF′′(z)), (3.101)

2G(ux + iuy) = κF(z) + F(z) − 2iyF′(z). (3.102)

From (3.101) we have at y = 0

τxy = −2ImF′(xo)

that is twice τy-value in the problem of the longitudinal shear in (3.5) and wecan replace in the results of sub-chapter 3.1 w′(z) by 2F′(z).

3.2.11 Rupture Due to Crack in Transversal Shear

In this case (Fig. 3.20) τxy(∞) = τ and we derive from (3.14)

F′(z) = −iτz/2√

z2 − l2,F(z) = −0.5iτ√

z2 − l2 (3.103)

and according to (3.102) we find on axis x at x < /l/ and x > /l/ respectively

τxy = uy = σy = 0, σx = −2τx/√

l2 − x2,ux = −τ((κ + 1)/2G)√

l2 − x2,

σx = σy = ux = 0, τxy = τx/√

x2 − l2,uy = ((κ − 1)/2G)√

x2 − l2

(3.104)

y

x

τ

τ

τ

τ

Fig. 3.20. Crack in transversal shear


and by the Clapeyron’s theorem (3.17) as well as the energy balance

(1 + κ)πlτ∗2dl/4G = 4γsdl

we compute

τ∗ = 4√

γs/π(κ + 1)l. (3.105)

The same results can be received according to the asymptotic approachand (3.102), (3.103) as

σr = −(K2/√

2πr)(2 − 3 cos2(θ/2)) sin(θ/2), σθ = −3(K2/√

2πr) cos2(θ/2) sin(θ/2),

τrθ = (K2/√

2rπ)(1 − 3 sin2(θ/2)) cos(θ/2), τe = (K2/2√

2πr)√

1 + 3 cos2 θ,(3.106)

uruθ = (K2/2G)√

r/2πx(−κ + 5 − 6 sin2(θ/2)) sin(θ/2)(−κ − 5 + 6 cos2(θ/2)) cos(θ/2).

(3.107)

Here K2 = τ√

πl – the stress intensity coefficient of the second crack task.Further computation follows that one for a crack in tension and we find asimilar value (see also (3.93))

K2∗ = 4√

Gγs/(κ + 1) (3.108)

and the strength conditionK2 ≤ K2∗ .

Diagrams σθ/σyi, σr/σyi, τ/σyi at τe = σyi/2 as functions of θ are givenin Fig. 3.21 by solid, broken and interrupted by points lines 1 respectively.

3.2.12 Constant Displacement at Transversal Shear

Using the analogy mentioned above we have from (3.8) at ζ = z

F′(z) = −Q/2π√

z2 − l2,F(z) = 2Guo− (Q/2π)ln(z/l+√

(z/l)2 − 1) (3.109)

0

0.5

−0.5

−1

0

0 30 60 90 120 150 θo

σθ/σyiσr/σyi

τ/σyi

0

0

0

0

0

1

11

1

1

Fig. 3.21. Diagrams of stress distribution at crack ends in transversal shear


wherein Q is a resultant of τxy at y = 0,−l < x < l. From (3.101), (3.102),(3.109) we find on axis x at x < /l/ and x > /l/ respectively

ux = uo,uy = σx = σy = 0, τxy = Q/π√

l2 − x2,

τxy = uy = σy = 0, σx = −2Q/π√

x2 − l2,

ux = uo − (Q/2πG)(κ + 1) ln(x/l +√

(x/l)2 − 1).

In the asymptotic approach we receive similarly to (3.106), (3.107) as

σr = −(Q/π√

2rl)(3 cos2(θ/2) − 1) cos(θ/2), σθ = −3(Q/π√

2rl) sin2(θ/2) cos(θ/2),

τrθ = (Q/π√

2rl)(3 sin2(θ/2) − 2) sin(θ/2), τe = (Q/2π√

2rl)√

1 + 3 cos2 θ, (3.110)

ur = u1 − (Q/πG)√

r/2l((κ + 1)/2 − sin2(θ/2)) cos(θ/2),

uθ = u2 − (Q/πG)√

r/2l((κ − 1)/2 − 3 cos2(θ/2)) sin(θ/2).

And we again can see that τe in (3.106), (3.110) differ by a constant multiplier.

3.2.13 Inclined Crack in Tension

By a combination of the solutions in Sects. 3.2.7, 3.2.11 a strength of a bodywith inclined crack in tension (Fig. 3.22) can be studied. Supposing accordingto (2.72) σ = p sin2 β, τ = 0.5p sin 2β and seeking in the end of the crack mainplane with θ = θ∗ L. Kachanov found in /17/ relation

sin θ∗ + (3 cos θ∗ − 1)cotβ = 0 (3.111)

according to which the crack must propagate in this direction. Some experi-ments confirm it.

y

p

p

x

θ*β

Fig. 3.22. Inclined crack in tension


3.3 Axisymmetric Problem and its Generalization

3.3.1 Sphere, Cylinder and Cone Under External and InternalPressure

For a sphere with internal a, current ρ and external b radii (Fig. 3.23) we useequations (2.80) and the Hooke’s law (2.17) at σθ = σχ in the form

σθ = E(εθ +νερ)/(1–ν−2ν2), σρ = E(ερ(1−ν)+2νεθ)/(1−ν−2ν2). (3.112)

Putting (2.81) into (3.112) and the result – in (2.80) we get on a differentialequation for uρ ≡ u

d2u/dρ2 + 2du/ρdρ–2u/ρ2 = 0

with an obvious integralu = A/ρ2 + Bρ. (3.113)

Now we determine strains from (2.80), stresses – by (3.112) and constants –according to boundary conditions σρ(a) = −q, σρ(b) = −p. As a result wehave

σθ = (qa3(2ρ3 + b3)–pb3(2ρ3 + a3))/2ρ3(b3a3),

σρ = (qa3(ρ3–b3) + pb3(a3 − ρ3))/ρ3(b3 − a3).(3.114)

The strains and the displacements can be found according to the Hooke’s lawand expressions (2.80).

In a similar way the stress distribution in a tube can be analysed. Tochange the method we use here potential function Φ (see Sect. 2.4.3) since theproblem is a plane one. The biharmonic equation (2.74) in this case becomes

d(rd(d(rdΦ/dr))/rdr)/dr)/rdr = 0

a

bq

pρ

Fig. 3.23. Sphere under internal and external pressure

3.3 Axisymmetric Problem and its Generalization 73

p

λ

χρ

Ψ

q

O

Fig. 3.24. Cone under external and internal pressure

with a very simple solution that with a help of (2.75) and boundary conditionslike that for the sphere (with replacement in them ρ by r) gives

σr = (a2b2(p − q)/r2 + qa2 − pb2)/(b2 − a2),

σθ = (qa2 − pb2 + a2b2(q − p)/r2)/(b2 − a2).(3.115)

Let us now consider a cone (Fig. 3.24) for which we use spherical coordinates(Fig. 2.10) and supposition τρθ = τρχ = τχθ = ερ = γρθ = γρχ = γχθ = 0 (as in acylinder). Other components do not depend on ρ, θ. Above that uρ = uθ = 0and uχ = ρu(χ). In coordinates θ, χ the first equation (2.77) takes the form

dσχ/dχ + (σχ–σθ)cotχ = 0 (3.116)

and expressions (2.79) give

u = C/ sin χ, εθ = −εχ = Ccos χ/ sin2 χ. (3.117)

Now we use the Hooke’s law (2.17) at ν = 0.5 that leads to relation

σθ − σχ = 4GC cos χ/ sin2 χ

and from (3.116) – to

σχ = D − 2GC(cos χ/ sin2 χ + ln tan(χ/2)). (3.118)

Constants C, D have to be determined from border demands σχ(ψ) = −q,σχ(λ) = −p. As a result we derive finally

σθσr

= −q + (q − p)(cos ψ/ sin2 ψ ± cos χ/ sin2 χ − ln(tan(χ/2)/ tan(ψ/2)))/A


where

A = cos ψ/ sin2 ψ– cos λ/ sin2 λ + ln(tan(ψ/2)/ tan(λ/2)). (3.119)

From expression (3.117) we find deformations and displacement for incom-pressible body as follows

εθ = (p − q) cos χ/2GA sin2 χ = −εχ,u = (p − q)/2AG sin χ.

This solution can model a behaviour of a volcano. When ψ, λ, χ tend to zerowe get the Lame’s relations for the tube that were derived above.

The theory of this section can be used for an appreciation of the strengthof different voids in a medium.

3.3.2 Boussinesq’s Problem and its Generalization

Stresses in Semi-space Under Concentrated Load

If an external concentrated force F acts vertically in point O (Fig. 2.10) on asemi-infinite solid the stresses in point N are /5/

σz = −3Fz3/2πρ5, σr = F((1 − 2ν)(ρ−z)/ρr2 − 3r2z/ρ5)/2π,

σθ = F(1 − 2ν)(zr2 + zρ2 − ρ3)/2πr2ρ3, τrz = −3Frz2/2πρ5.

(3.120)

These relations are known as Bousinesq’s solution for axisymmetric problempublished in 1889 and they are similar to Flamant’s expressions in Sect. 3.2.3for plane one. Using (2.72) we compute

σρ = F((1 − 2ν)(1 − z/ρ) − 3z/ρ)/2πρ2,

σχ = Fz2(1 − 2ν)(1 − z/ρ)/2πr2ρ2,

τρχ = Fz(1 − 2ν)/2πrρ2

(3.121)

and we can see that only for incompressible material (ν = 0.5) directions ρ, χare main ones and σχ = σθ = 0.

Stresses Under Distributed Load

Using the superposition method we can find stresses under any load. As thefirst example we consider a circle of radius a under uniformly distributed forcesq. Firstly we study stresses along axis z where we have /5/

σz = q(z3(a2 + z2)−3/2–1). (3.122)

In the same manner stresses σr, σθ (Fig. 3.25) can be found as

σθ = σr = q(−1− 2ν + 2(1 + ν)z/√

a2 + z2 − 3z3(a2 + z2)−3/2/2)/2. (3.123)


a ar

p

q

o

z

σθ

σr

σr

σZ

σZ

Fig. 3.25. Stresses under uniformly distributed load in circle

Particularly in point O we have

σz = −q, σr = σθ = −q(1 + 2ν)/2.

The maximum shearing stress can be easily computed according to (2.10),(3.122), (3.123) as follows

τe = q(0.5(1 − 2ν) + (1 + ν)z/√

a2 + z2 − 3z3(a2 + z2)−3/2)/2. (3.124)

This expression has its maximum at z∗ = a√

(1 + ν)(7 − 2ν) and it is

max τe = q(0.5(1 − 2ν) + 2(1 + ν)√

2(1 + ν)/g)/2. (3.125)

For example if ν = 0.3 then z∗ = 0.64a and max τe = 0.33q.An interesting case takes place for a circular punch and Boussinesq gave

the solution in a form similar to (3.95) as

q = P/2πa√

a2 − r2 (3.126)

where P is a resultant of loads q. The least value of q is in the centre: qmin =P/2πa2. Diagram q(r) is given in Fig. 3.26 by broken line and as we can seethe stresses are very high at r = a (similar to other problems of punchesand cracks in plane problem). In reality plastic strains appear at the edges,redistribution of stresses occurs and q(r) diagram has a form of the solid curvein the figure.

Stresses Under Rectangles

The linear dependence of stresses on displacements allows to use the super-position principle for finding stresses at different loadings. To realize that we


aZ

P

r

Fig. 3.26. Distribution of stresses under circular punch

rewrite the first relation (3.120) for the stress in a point with coordinates z, r(Fig. 2.10) as

σz = KσF/z2. (3.127)

Here (in this section compressive stresses are taken positive)

Kσ = 3/2π(1 + r2/z2)5/2

is a coefficient the values of which are given in special tables (see Appendix B).When several (n) forces act then stress σz is computed as follows

σz =

(n∑

i=1

KσiFi

)/z2

where factors Kσi are taken as the functions of ratio ri/z and ri is the distancefrom the studied point to the direction of a Fi action. This method can beapplied to a case of distributed load when we lay out a considered area onseparate parts and compute the resultant for each of them.

The special particularly important case takes place when we have uni-formly distributed load over a rectangle. Here we lay out the whole area onseparate rectangles and find the stress in the common for them point as asum of the stresses in each of the parts. The following options can be met(Fig. 3.27):

1) point M is on a border of the rectangle (Fig. 3.27, a) and we summarizestresses due to loads in rectangles abeM and Mecd,

2) point M is inside a rectangle (Fig. 3.27, b) and we summarize the stressesfrom the action of the load in rectangles Mhbe, Mgah, Mecf and Mfdg,

3) point M is outside a rectangle (Fig. 3.27, c) and we summarize the stressesfrom the action of a load in rectangles Mhbe and Mecf and subtract thatin rectangles Mhag and Mgdf.


b

b

h

e

e c b

f

e c

c

dM

M

M

c)b)

a)

f

a

a g

gd

d h

a

I II

Fig. 3.27. Uniformly distributed load over rectangle

The determination of stresses is fulfilled with the help of special tablesaccording to relation

σz = K′q (3.128)

where factor K′ is given in the function of ratios m = l/b – relative lengthand n = z/b – relative depth (see Appendix C). q is an intensity of the loads.E.g. for the case 1) we have

σz = q((K′)I + (K′)II). (3.129)

Displacements in a Massif

We begin with the case of concentrated force F when we have according tothe Hooke’s law on the surface z = 0 /5/

ur = −(1 − 2ν)(1 + ν)F/2πEr,uz ≡ S = F(1 − ν2)/πEr. (3.130)

In other cases we use the superposition method. E.g. for a circle of radius aunder uniformly distributed load q we write for a point outside it

uz = q(1 − ν2)r(L(a/r) − (1 − a2/r2)K(a/r))/πE (3.131)

where K(a/r), F(a/r) are full elliptic integrals of the first and the second kind.They can be calculated with a help of special tables. For the settling of theexternal circumference (r = a) we receive

uz = 4(1 − ν2)qa/πE (3.132)

and in points inside the circle the displacement is

uz = 4(1 − ν2)qaL(a/r)/πE. (3.133)


The highest displacement is in the centre of the circle as

max uz = 2(1 − ν2)qa/E

and it is easy to prove that max uz/uz(a) = π/2. Now we find a meandisplacement as

meanuz = (P/πa2)

a∫

0

2πuzrdr = 0.54P(1 − ν2)/πE

and it is near to the displacement under a circular punch

uz = 0.5P(1 − ν2)/πE. (3.134)

A similar situation takes place for a square with sides 2a loaded byuniformly distributed forces q. In this case

max uz = 8qaln(√

2 + 1)(1 − ν2)/πE = 2.24qa(1 − ν2)/E. (3.135)

In corners uz = 0.5max uz and an average uz is equal to 1.9qa(1−ν2)/E. Thesame computations were made for rectangles with different ratios of h/b. Theresults are represented in a form

uz = moq(1 − ν2)/E. (3.136)

The values of mo are given in table here as a function of the sides ratio h/b.

h/b circle 1 1.5 2 3 5 10 100mo 0.96 0.95 0.94 0.92 0.88 0.82 0.71 0.37

Approximate Methods of Settling Computations

In practice some approximate approaches are used for a computation of asettling. One of them is a method of a summation “layer by layer”. Herethe hypothesis is taken that a lateral expansion is absent or in other wordsthat the dependence of stresses on porosity is compressive (see Sect. 1.4.2). Itis also supposed that a decrease of σz with a depth subdues to Boussinesq’ssolution (Fig. 3.28). The whole settling is calculated as a sum of displacementsof elementary layers /10/

S = βo

n∑i=0

σzihi/Ei. (3.137)

Here βo is a dimensionless coefficient equal usually to 0.8, hi, Ei – a thicknessand a modulus of deformation of i-layer, σzi is computed according to the firstrelation (3.120) for the middle of the layer, hn is taken for a layer where thesettling is small. In Russia σzn = 0.2σze where σze are stresses from earth’s


σze

σz1

σz2

σzi

σzn

0.2 σze

hn

z

bp

ha

hi

h1

h2

Fig. 3.28. Approximate computation of settling

self-weight (see Sect. 2.4.1). If this layer has E < 5MPa it is included in sum(3.137). For hydro-technical structures with big width b (Fig. 3.28) conditionσz > 0.5σze is usually taken.

Another approach to the solution of this problem gave N. Cytovich /3/who proposed to take into account some lateral expansion of the soil and aninfluence of a footing size (see Fig. 1.6). He introduced the so-called equiva-lent layer hs which exposes the same settling as in the presence of a lateralexpansion:

hs = (1 − ν2)ηb. (3.138)

Here parameter η considers a form and a rigidity of a footing with width b.When a foundation has a form of a rectangle the method of corner points

is applied similar to that for the calculation of stresses.

3.3.3 Short Information on Bending of Thin Plates

General Equations for Circular Plates

A plate is considered to be thin when the ratio of its thickness to the minimumdimension in plane L satisfies the condition 0.2 > h/L > 0.0125. This problemis studied in special courses and comparatively simple theory exists for axi-symmetric plates. Differential equation of their element (Fig. 3.29) is

Mθ − d(Mrr)/dr = Qr. (3.139)

Here Mr, Mθ are radial and tangential bending moments, Q – transversal(shearing) force which can be computed according to an equilibrium conditionof a middle part of the plate with radius r. In the case of uniformly distributedload q it is

Q = 0.5qr. (3.140)


Q

Q

dr

r

dθ

Mθdr

Mθdr

Mrrdθ

(Mrr+d(Mrr))dθ

Fig. 3.29. Element of circular plate

For a plate of radius R at q = constant in a circle of diameter 2ro integra-tion of (3.139) with consideration of (3.140) gives equation

RMr(R) + q(ro)3/6 + q(ro)2(R − ro)/2 =

R∫

0

Mθdr. (3.141)

Replacing in (3.141) q(ro)2 by F/π and supposing ro = 0 we come to the caseof concentrated force F in the centre of the plate as

FR/2π =

R∫

0

Mθdr − RMr(R). (3.142)

At R = ro we have from (3.141) the solution for a plate under uniformlydistributed pressure q in form

RMr(r) + qR3/6 =

R∫

0

Mθdr. (3.143)

Similarly some other cases can be considered.

Ultimate State of Circular Plates

Now we find according to the first Gvozdev’s theorem the ultimate state ofthe plate. If its edges are freely supported we must put in (3.142), (3.143)Mr(R) = 0, Mθ = M∗ (where M∗ = σyih2/4 – see expression (1.24)) whichgives

F∗ = 2πM∗, q∗ = 6M∗/R2. (3.144)


Comparing these F∗ and q∗ to (1.28) and the first (1.31) we see that theycoincide and hence are rigorous. If the edges of the plate are fixed we mustput in (3.142), (3.143) −Mr(R) = Mθ = M∗. That leads to ultimate values

F∗ = 4πM∗, q∗ = 12M∗/R2 (3.145)

which coincide with relation (1.30) and the second expression (1.31) respec-tively. Therefore they are also exact. In the similar way some other differentcases of an axi-symmetric load can be considered.

Ultimate State of Square Plates

The bending of rectangular plates are usually studied in double trigonomet-ric series. E.g. for a square 2Rx2R in plane loaded by uniformly distributedpressure q with origin of coordinate system x, y in one of its corners we have

MxMy

= (64qR2/π4)∞∑

m=1 n=1

∞m+νn2∑n+νm2

(x(sin mπx/2R)

(sin nπy/2R))/mn(m2 + n2)2(m,n = 1, 3, ...). (3.146)

Taking only the first member of the series we find for maximum moments (inthe centre of the plate)

max Mx = max My = (1 + ν)qR2/6

which give the ultimate load as

q∗ = 6M∗/(1 + ν)R2. (3.147)

Another simple solution can be received when we use differential equationof an element of the plate in form

∂2Mx/∂x2 + 2∂2Mxy/∂x∂y + ∂2My/∂y2 = −q (3.148)

where Mxy is the moment of a torsion. Taking approximately for the momentsexpressions that satisfy the border demands (here the origin of the coordinatesystem is in the centre of the plate) Mx = C(R2 − x2), My = C(R2 − y2),Mxy = 0 and putting them into (3.148) we find C = q/4 and hence

q∗ = 4M∗/R2

which coincides with (3.147) for incompressible material.Taking into account (3.147) and the first relation (1.31) we find the

following limits for the ultimate load

6M∗/R2(1 + ν) ≤ q∗ ≤ 6M∗/R2. (3.149)

We can see that the q∗-value is rigorous for ν = 0.


p p

p

a

r

z

p

p

p

p

p

Fig. 3.30. Circular crack in tension

3.3.4 Circular Crack in Tension

Here (Fig. 3.30) equations (2.76) are valid and we seek solution in the followingform /17/

uz = −∂2Φ/2G∂r∂z,uz = (2(1 − ν)∆Φ − ∂2Φ/∂z2)/2G

where Φ is a function of z and r. Putting these expressions into (2.76) andthen strains – into the Hooke’s law we get the stresses as

σr = ∂(ν∆Φ − ∂2Φ/∂r2)/∂z, σθ = ∂(ν∆Φ − ∂Φ/r∂r)∂z,σz = ∂((2 − ν)∆Φ − ∂2Φ/∂z2)∂z, τrz = ∂((1 − ν)∆Φ − ∂2Φ/∂z2)/∂r.

These values satisfy static equations and putting them into compatibilityrelation (2.76) we find biharmonic law for Φ.

Using the Henkel’s transformations we get expressions for the displacementand stress that at p = constant, ρ = r/a and z = 0 are

uz(ρ, 0) = 4(1 − ν2)pa√

1 − ρ2/πE (ρ ≤ 1),σz = 2p(1/

√ρ2 − 1 − sin−1(1/ρ))/π (ρ > 1).

(3.150)

Since near the crack edges the first member in brackets is much higher thanthe second one the solution is somewhat similar to that (3.126) for the circularpunch.

From Fig. 3.31 where the curve uz(ρ) according to the first (3.150) is shownby broken lines we can see that deformed crack is an ellipsoid. Stress σz hasthe same peculiarity as in similar problems at the longitudinal shear and planedeformation. Using the expression for the work at crack propagation

W = 2pπa2

1∫

0

uz(ρ, 0)ρdρ,


a a

uz(r,0)

r

z

Fig. 3.31. Deformation of crack

relation (3.150) for uz(ρ, 0) and equality

dW = 2πγsada

as in Sects. 3.1.4, 3.2.8 and 3.2.11 we find

Ko∗ =√

γsE/(1 − ν2) = 2p√

a/√

π.

The problem of non-uniformly distributed forces is solved in the same manner.The task of two cracks in distance z = ±z1 is also studied in a space and ina cylinder.

4

Elastic-Plastic and Ultimate State of PerfectPlastic Bodies

4.1 Anti-Plane Deformation

4.1.1 Ultimate State at Torsion

Although exact elastic solutions at torsion are known only for somecross-sections the ultimate state can be found for any problem because inthis case we should consider only two equations for two unknowns (Fig. 2.8) –condition τe = τyi together with static law (2.48). It can be satisfied if wetake

τx = ∂w/∂y, τy = ∂w/∂x (4.1)

and from (2.52) we find

(∂w/∂x)2 + (∂w/∂y)2 = (τyi)2

or/gradw/ = τyi = constant. (4.2)

Here the gradient w(x, y) is the maximum slope of that function which canbe interpreted as a sand heap with angle of repose equal to tan−1 τyi. Expres-sion (4.2) means that the distance between lines in which τyi acts are constantand it allows to compute an elementary moment of torsion as (Fig. 4.1)

dM∗ = τyipdpds = τyi2dpdA

where A is an area under curve w = constant and p–perpendicular to it froma pole. Summarizing dM∗ we find the ultimate moment as

M∗ = 2V. (4.3)

Here V is volume of the heap.Relation (4.3) opens the way for finding the ultimate load experimentally.

For some sections M∗-value can be calculated. In the case of a circle withradius R e.g. we have

86 4 Elastic-Plastic and Ultimate State of Perfect Plastic Bodies

ds

dp

pA

τyi

Fig. 4.1. Computation of torsion moment

b

h

Fig. 4.2. Ultimate moment for rectangle

M∗ = 2πR3τyi/3.

For a rectangle (Fig. 4.2) we compute in a similar way

M∗ = h2(3b − h)τyi/6.

From this expression we have as particular cases M∗-values for a long stripand a square:

M∗ = bh2τyi/2,M∗ = h3τyi/3.

Relations above can be used for τyi determination by the torsion tests ofplastic materials including some soils.

4.1.2 Plastic Zones near Crack and Punch Ends

When plastic strains appear the value of τ∗ (see Sect. 3.1.4) falls as G decreasesin many times. There are several solutions for a perfect plastic body. Wefollow here the approaches of Rice /21/ for small plastic zones where relations(3.19) are valid. As we mentioned there the condition τe = constant gives acircumference and we can suppose that the plastic zone is a circle (Fig. 4.3)with a radius Ro that can be found from compatibility conditions for stressesand displacement uz on the border between elastic and plastic districts.

4.1 Anti-Plane Deformation 87

OO1

r1

r

Ro

Fig. 4.3. Plastic zone near crack end

According to relations of (2.55) type we find with a help of (3.19)

τr1 = 0, τθ1 = τo

√l/2r.

Taking τe as τyi we receive Ro as follows

Ro = (τo)2l/2(τyi)2. (4.4)

Now we consider the displacements and from elastic part of the body we haveaccording to (4.4) and the first expression (3.19)

uz = (τo/G)√

2lRo sin θ/2. (4.5)

Then from the second expression (2.57) we find on the circumference start-ing from the plastic zone at r1 = 2Ro cos θ1

uz = (τyi/G)2Ro

θ1∫

0

cos θ1dθ1

or after computationsuz = 2(τyi/G)Ro sin θ1. (4.6)

Taking into account equality θ1 = θ/2 and comparing (4.6) to (4.5) we geton expression (4.4) that is the same value of Ro. Lastly we determine thedisplacement in the end of the crack in form

δ = 2uz(Ro, π) = 2l(τo)2/Gτyi.

In the same manner the problem of the strip’s longitudinal movement canbe studied. Using expressions (2.58) and (3.10) we find on the circumferencer = Ro:

τθ1 = 0, τr1 = τe = Q/π√

2Rol.

Supposing τr1 = τyi we find the radius Ro of the plastic zone as

Ro = Q2/2π2l(τyi)2. (4.7)


Now if we replace in the third relation (3.10) r by Ro this equation isvalid on the border from the elastic side. And from the plastic region we haveaccording to the first (2.57)

uz = uo − τyir1/G oruz = uo − 2Ro(τyi/G) cos θ/2.

Comparing this expression to the first relation (3.10) at r = Ro we find again(4.7). The position of the circumference’s centre will be determined in thenext chapter.


4.2.1 Elastic-Plastic Deformation and Failure of Slope

Stresses in Wedge

As was told in Sect. 3.2.1 the maximum shearing stress τe in the cases λ > π/4reaches its maximum at θ = 0 and there the first residual strains appear whenload p is

pyi = 2τyi(2λ cos λ − sin 2λ)/(cos 2λ − 1).

At p > pyi we have in Fig. 3.5 plastic zone BOC as well as two elasticdistricts AOB and COD for which the solution of Sect. 3.2.1 is valid in form

τ = C1 + C2 cos 2θ + C3 sin 2θ, (4.8)σθσr

= C4 − 2C1θ ± C3 cos 2θ ± (−C2 sin 2θ).

In the plastic zone we take the same condition as on the straight line θ = 0 atp = pyi that is τ = τyi and from (3.21) we find with consideration of demandσr(0) = σθ(0) = −p/2

σr = σθ = −p/2 − 2τyiθ.

Constants Ci(i = 1. . .4) and τyi should be found from the compatibility equa-tions for stresses at θ = ± υ and boundary conditions τ(−λ) = σθ(−λ) = 0and τ(λ) = 0, σθ(λ) = −p on lines OA and OD respectively. As a result wehave in the plastic zone

τ = Co(cos 2(λ − υ) − 1),σr = σθ = 2Coθ(1 − cos 2(λ − υ)) − p/2

(4.9)

whereCo = 0.5p/(−2υ + 2λ cos 2(λ − υ) − sin 2(λ − υ)) (4.10)

and in elastic districts AOB, COD at upper and lower signs before υ,respectively


τ = Co(cos 2(λ − υ) − cos(θ ± υ)),

σθσr

= Co(±2υ − 2θ cos 2(λ − υ) ± sin 2(θ − (±υ)) − p/2.

At λ−υ = π/4 and τ = τyi we have from (4.9), (4.10) the ultimate load as

pu = 2τyi(2λ − π/2 + 1) (4.11)

and it is interesting to notice that if we take the solution that is recommendedin /18/ by V. Sokolovski for the case λ ≤ π/4 which in our case gives thesmaller load at π/2 > λ > π/4 as follows

(pu)′ = 2τyi(sin 2λ − (π/2 − 1) cos 2λ).

However the last relation predicts a fall of the ultimate load with an increaseof λ as a whole (e.g. (pu)′(π/2) = 1.14τyi) that contradicts a real behaviourof foundations.

Displacements in Wedge

In order to find displacements we use expressions (2.69), (2.66) in which m = 1,Ω = 1/G and indices x, y are replaced by r, θ, respectively. As a result wehave in districts AOB and COD at upper and lower signs consequently

ur = D1 cos θ + D2 sin θ + (Cor/2G) sin 2(θ − (±υ)),uθ = −D1 sin θ + D2 cos θ + (Cor/2G)(D3 + cos 2(θ − (±υ)) − 2(lnr) cos 2(λ − υ)).

Here D1, D2, D3 should be searched from compatibility equations at θ = υ.An anti-symmetry demand gives D1 = 0. At ultimate state we have the dis-placements of lines AO, OD as

uθ = ±(−D2 cos λ) + D3Cor/2G

and since the movement in infinity must have finite values we should putD3 = 0. So the solution predicts parallel transition of lines OA, OD (brokenlines in Fig. 3.5).

Ultimate State of Slope

As an alternative we study a possibility of a rupture in the plastic zone whereelongations ε1 = γ/2 take place. From expression (2.32) we write

τ = 2G(t)ε1exp(−αε1)

and according to criterion dε1/dt → ∞ we find the critical values of γ and tas follows

ε∗ = 1/α,G(t∗) = pαe(cos 2(λ−υ)− 1)/4(2λ cos 2(λ−υ)− 2υ− sin 2(λ−υ)).

If the influence of time is negligible the ultimate load can be determined as

p∗ = 4G(2λ cos 2(λ − υ) − 2υ − sin 2(λ − υ))/αe(cos 2(λ − υ) − 1).

The smallest value of p∗ and pu (see relation (4.11)) must be chosen.


4.2.2 Compression of Massif by Inclined Rigid Plates

Main Equations

Here we use the scheme in Fig. 3.6. Excluding from (2.65), (2.68) at τe = τyi

difference σr − σθ we get on an equation for τrθ ≡ τ at τ = τ(θ) which afterthe integration becomes

dτ/dθ = ±(−2√

(τyi)2 − τ2) + 2nτyi (4.12)

where n is a constant. The integration of (4.12) gives a row of useful results.When n = 0 we find expression τ = ±τyi sin(c + 2θ) which corresponds

to homogeneous tension or compression. The family of these straight lineshas two limiting ones on which τ = ±τyi (they are called “slip lines”) andaccording to the first two equations (3.21) σr = σθ = ±2τyiθ. Another familyof slip curves is a set of circular arcs (Fig. 4.4, a), Such a field was realized inplastic zone BOC of the problem in Sect. 4.2.1 and can be seen near punchedges. The photographs of compressed marble and rock specimens are givenin book /22/ and they are shown schematically in Fig. 4.4, b. It is interestingto notice that this stress state is described by the same potential function(see (2.75))

Φ = τyir2θ

as in an elastic range.

Common Case

When in (4.12) n = 0 we have a compression of a wedge by rough rigid plates.Putting in (4.12)

τ = τe sin 2ψ, σr − σθ = 2τe cos 2ψ (4.13)

a) b)

P

Fig. 4.4. Slip lines


10

40

80

120

λo

2 3 4 n

Fig. 4.5. Dependence λ on n

where ψ is equal to angle Ψ in Figs. 1.21 and 1.22 we find for the upper signin (4.12)

dψ/dθ = n/ cos 2ψ − 1. (4.14)

The integral of (4.14) at boundary condition ψ(0) = 0 is obvious

θ = n(n2 − 1)−1/2 tan−1(√

(n + 1)/(n − 1) tan ψ) − ψ

and n depends on λ according to the second border demand ψ(λ) = π/4as (Fig. 4.5)

λ = n(n2 − 1)−1/2 tan−1√

(n + 1)/(n − 1) − π/4.

Now from static equations (2.67) we computeσrσθ

= τyi(C − 2nln(r/a) − nln(n − cos 2ψ) ± cos 2ψ)

where constant C can be found from the first equation (3.32). The simplestoption is

σrσθ

= τyi(2nln(r/a) − nln((n − cos 2ψ)/(n − 1)) ± cos 2ψ). (4.15)

Sokolovski /18/ used this solution for the description of material flow througha narrowing channel. For this case we can find resultant Q = ql (Fig. 3.6)according to the second integral static equation (3.32) as /23/

q = 2nτyi((a/l + 1)ln(l/a + 1) + 0.5 ln(n/(n − 1)) − 1).

Diagrams σθ(r/a) and τe(λ) are given by pointed lines in Figs. (3.6). . .(3.9).We can see that the distribution of σθ is more uneven and τe = τyi is muchsmaller than according to the elastic solution.

In order to find displacements we use relations (2.69) which give

ur = uo/r(n − cos 2ψ) − Vo cos θ/ cos λ,uθ = Vo sin θ/ sin λ

where Vo is the plates displacement and uo is unknown. It should be foundfrom an additional condition.


Cases of Big n and Parallel Plates

If n is high we have from (4.14)

dψ/dθ = n/ cos 2ψ

and after integrationnθ = 0.5 cos 2ψ

Parameter n is linked with λ as n = 1/2λ and for ψ we have

sin 2ψ = θ/λ, cos 2ψ =√

1 − (θ/λ)2.

In the same manner as before we find stresses and displacements

τ = τyiθ/λ, σrσθ

= τyi(λ−1 ln(a/r) − 1 +√

1−(θ/λ)2

0 ),

uθ = Vo sin θ/ sin λ,ur = uo

√λ2 − θ2 − Vo cos θ/ sin λ.

Lastly at λ → 0 we have the case of parallel plates and at y = aθ,h = aλ(Fig. 4.6),

λ−1ln(r/a) = x/h

τ = τyiy/h, σy = −τyi(1 + x/h), σx = −τyi(1 + x/h − 2√

1 − (y/h)2). (4.16)

From integral static equation we compute

p = P/l = τyi(1 + l/2h).

Diagrams σx(y) and σy(x) for the left side of the layer are shown in Fig. 4.6.The broken lines correspond to the case when the material is pressed intospace between the plates (two similar states are described in Sect. 1.5.4). Inorder to find displacements we suppose uθ = −Voy/h and according to (2.60)we compute

x

σx

y

l

h

h

Fig. 4.6. Compression of massif by parallel plates


ux = Vo(x/h + 2√

1 − (y/h)2).

The set of slip lines is also drawn in Fig. 4.6.They are cycloids and theirequation will be given later. Experimental investigations show that rigid zonesappear near the centre of the plate (shaded districts in Fig. 4.6) while plasticmaterial is pressed out according to the solution (4.16) above.

Its analysis shows that at small h/l shearing stresses are much less thanthe normal ones and the material is in a state near to a triple equal tension orcompression. This circumstance has a big practical and theoretical meaning.It explains particularly the high strength of layers with low resistance to shearin tension (solder, glue etc.) or compression (soft material between hard onein nature or artificial structures). It also opens the way to applied theory ofplasticity /10/.

Addition of Shearing Force

Here we suppose /10/ that shearing stresses on contact surfaces (Fig. 4.7) areconstant. At y = h, x < l and y = −h, x > l we have τ = τyi and in otherparts of the surface τ = τ1 < τyi. Then satisfying static equations (2.59) andcondition τe = τyi the solution may be represented in a form

τxy/τyi = (1 + k1)/2 + (1 − k1)y/2h,

σy/τyi = −C − (1 − k1)x/2h, σx/τyi = σy/τyi + 2√

1 − (τxy/τyi)2. (4.17)

Here k1 = τ1/τyi and C is a constant. If k = −1 we have solution (4.16) andat k = 1 we receive a pure shear (σx = σy = 0, τxy = τyi).

Now we use integral static equations similar to (3.32)

h∫

−h

σx(0, y)dy = 0,

1∫

0

σy(h)dx = p

ll 2P

2P

2Q

yh

h

x

τ1

τ1

τyi

τyi

2Q

Fig. 4.7. Layer under compression and shear


where p = P/lτyi which give after exclusion of C

π/2 − k1

√1 − (k1)2 − sin−1 k1 = (1 − k1)(−p − (1 − k1)l/4h). (4.18)

Then we take integral equilibrium equation at contact surface as

2Q = τyi(1 + k1)l

which gives 1+k1 = 2q where q = Q/τyil. Excluding from (4.18) k1 we finallyreceive

(1−q)(−2p− (1−q)l/h) = π/2+2(1−2q)√

q(1 − q)− sin−1(2q−1). (4.19)

At q = 0 we again find Prandtl’s solution (4.16).From Fig. 4.8 where diagrams (4.19) for l/h = 10 and l/h = 20 are

constructed we can see the high influence of q on ultimate pressure p.

4.2.3 Penetration of Wedge and Load-Bearing Capacityof Piles Sheet

As we can see from Fig. 4.5 the dependence λ(n) may be also used at λ >π/2 when a wedge penetrates into a medium (Fig. 4.9). General relations forstresses of Sect. 4.2.2 are valid here but constant C should be searched fromequations similar to (3.32) as

p∗ sin λ =

λ∫

0

(σr(a, θ) cos θ + τ(a, θ) sin θ)dθ,

P∗ = 2

⎛⎝p∗b +

a+1∫

a

(σθ(r, λ) sin λ + τ(r, λ) cos λ)dr

⎞⎠ . (4.20)

0 6 p

q

0.5l/h = 20

l/h = 10

Fig. 4.8. Dependence of p on q


P

bbc

l

a

r

*

p*

Fig. 4.9. Penetration of wedge

where p∗ is an ultimate pressure at compression. Putting into (4.20) σr, σθfrom (4.15) and τ from (4.13) we find

P∗/2lτyi = p∗(b/l+sin λ)/τyi−Jo−n(lnn−2+2(1+a/l)ln(l/a+1) sin λ)+cos λ.(4.21)

Here

Jo =

λ∫

0

(cos 2ψ − nln(n − cos 2ψ) cos θ + sin 2ψ sin θ)dθ.

In the case of a wedge penetration we must put in (4.21) a = 0 that givesthe infinite ultimate load due to the hypothesis of constant form and volumeof the material near the wedge. Because of that we recommend for the casethe solution of Sect. 4.2.2. However for λ near to π (an option of pile sheet)simple engineering relation can be derived when at n = .07, λ = 179, a → ∞we have from (4.21)

P∗ = 2(p∗b + τyil(1 + Jo)). (4.22)

The computations of Jo(π) gives its value 1.13. Taking into account the struc-ture of (4.22) and its original form (4.20) we can conclude that the influenceof σθ is somewhat higher than that of τ. We must also notice that P∗-value in(4.22) is computed in the safety side because we do not consider an influenceof σθ on τyi.

4.2.4 Theory of Slip Lines

Main Equations

Such rigorous results as in previous paragraphs are rare. More oftenapproximate solutions are derived according to the theory of slip lines that


can be observed on polished metal surfaces. They form two families of per-pendicular to each other lines for materials with τyi = constant. We denotethem as α, β and to find them we use transformation relations (2.72) whichgive the following stresses in directions inclined to main axes 1, 3 underangles π/4 (Fig. 4.10)

σα = σβ = σm = 0.5(σ1 + σ3),

ταβ = τyi = 0.5(σ1 − σ3).(4.23)

Now we find the stresses for a slip element in axes x, y. According to expres-sions (2.72) (Fig. 4.11)

σxσx

= σm ± τyi sin 2ψ, τxy = −τyi cos 2ψ. (4.24)

These relations allow to find equations of slip lines in form

Fig. 4.10. Stresses in element at ideal plasticity

Fig. 4.11. Slip element in axes x, y


dy/dx = tanψ = (1 − cos 2ψ)/ sin 2ψ = 2(τyi + τxy)/(σx − σy)

and for another family dy/dx = − cot ψ.

Examples of Slip Lines

Reminding the problem of the layer compression (see paragraph 4.2.2) we putin the last expressions the relations for stresses and get on equations

dy/dx = −√

(h − y)/(h + y),dy/dx =√

(h + y)/(h − y)

and after integration we find the both families of the slip lines as

x = C +√

h2 − y2 + hcos−1(y/h), x = C +√

h2 − y2 − hcos−1(y/h)

where C is a constant. The slip lines according to these expressions are shownin Fig. 4.6. In a similar way the construction of slip lines can be made for thecompressed wedge in Fig. 3.6.

As the second example we consider a tube with internal a and external bradii under internal pressure q. Here τrθ = 0, σr − σθ = 2τe = σyi and fromthe first static equation (2.67) we receive

q∗ = σyiln(b/a). (4.25)

Slip lines are inclined to axes r and θ by angle π/4 (broken lines in Fig. 4.12).From this figure we also find differential equation

dr/rdθ = ±1

with an obvious integralr = ro exp(±θ). (4.26)

So, the slip lines are logarithmic spirals which can be seen at pressing of asphere into an plastic material.

Fig. 4.12. Slip lines in tube under internal pressure


Construction of Slip Lines Fields

In order to construct a more general theory of slip lines we transform staticequations (2.59) into coordinates α, β putting there expressions (4.24). Apply-ing the method of Sect. 2.4.3 (see also /10/) we derive differential equations

∂(σm + 2τyiψ)/∂α = 0, ∂(σm + 2τyiψ)/∂β = 0

with obvious integrals

σm/2τyi ± ψ =ξη= constant. (4.27)

The latter formulae allows to determine ξ, η in a whole field if they areknown on some its parts particularly on borders. In practice simple con-structions are used corresponding as a rule to axial tension or compression(Fig. 4.13) and centroid one (Fig. 4.4, a). A choice between different optionsshould be made according to the Gvozdev’s theorems /9/.

Construction of Slip Fields for Soils

In a similar way the simple fields of slip lines can be found for a soil with angleof internal friction ϕ (see solid straight line in Fig. 1.22) when according to(1.34), (1.35) the slip planes in a homogeneous stress field are inclined to theplanes with maximum and minimum main stresses under angles π/4−ϕ/2 andπ/4 + ϕ/2, respectively. In order to generalize the centroid field in Fig. 4.4, awe find from Fig. 1.22 expression τ = ±(−σθ tan ϕ) and put it into the secondequation (3.21) which after transformations gives

σθ = Cexp(±2θ tan ϕ), τ = ±(−C(tan ϕ)exp(±2θ tan ϕ)).

Now we again use Fig. 4.22 and write the result at the upper signs in theprevious relations as follows

Fig. 4.13. Slip lines at homogeneous tension or compression


σm = σθ + τ tan ϕ = C(1 + tan2 ϕ) exp(2θ tan ϕ)

or finallyσm = Dexp(2θ tan ϕ) (4.28)

where D is a constant.Supposing that in the origin at r = ro the second family of the slip lines is

inclined to the first set of them (the rays starting from the centre – see Fig. 4.4)under angle π/4−ϕ/2 we conclude from Fig. 1.22 that they form angle ϕ withthe normal to r. So for the second family we have equation similar to the caseof τyi = constant as

dr/rdθ = tan ϕ

and hencer = ro exp(θ tan ϕ) (4.29)

(see also (4.26) and Fig. 4.12). This theory can be generalized for a cohesivesoil by the replacement in (4.28) σm by σm+c/ tan ϕ (broken line in Fig. 1.22).

4.2.5 Ultimate State of Some Plastic Bodies

Plate with Circular Hole at Tension or Compression

We begin with a simple example of a circular tunnel (Fig. 4.14) in a massifunder external homogeneous pressure p. In this case we choose a slip linesfield corresponding to simple compression (left side in the figure). Then wehave according to relations (4.27) σx = σm + τyi = 0 that means σm = −τyi

and σy = σm − τyi = −2τyi = −σyi. We suppose also that the material insidea strip 2a is rigid and we find

P∗ = 2(b − a)σyi. (4.30)

Since the P∗-value is found from the static equation the result is a rigorousone. It is also valid for a tension of the plane with a circular hole and it ismuch simpler than the similar solution for an elastic body in Sect. 3.2.5.

Penetration of Wedge

Now we consider a pressure of a wedge into a massif (Fig. 4.15). We supposethat a new surface OA is a plane and the slip field consists of two trianglesOAB, OCD at pure compression and a centroid part OBC between them.Firstly we determine the stress state in the triangles. In AOB

ψ = −υ/2, σ1 = σm + τyi

that means σm = −τyi. Similarly in COD

ψ = υ/2, σ3 = −p∗

that gives σm = τyi − p∗.


b P*

p*a a

b

y

x

Fig. 4.14. Compression of massif with circular tunnel

xO

G K

DC

B

y

A h

p*

EI

P*

Fig. 4.15. Penetration of wedge

Putting these results into (4.27) we receive

−τyi/2τyi − υ/2 = (τyi − p∗)/2τyi + υ/2

from whichp∗ = σyi(1 + υ) (4.31)

and according to static equation as the sum of the forces on vertical direction:

P∗ = 2σyi(1 + υ)lsinλ. (4.32)

The auxiliary quantity υ can be excluded by the condition of the equalityof volumes KDG and AOG. Since from triangle AOG angle OAG is equal toπ − (π/2− λ)− (π/2 + υ) or after cancellation - to λ − υ we have for segmentKE

lcosλ − h = lsin(λ − υ) (4.33)

and we find /24/


2

1

0

Fig. 4.16. Dependence of compressing force on angle λ

h2 tan λ = (lcosλ − h)(lcos(λ − υ) + (lcosλ − h) tan λ. (4.34)

Excluding from (4.33), (4.34) l, h we finally derive

2λ = υ + cos−1(tan(π/4 − υ/2)). (4.35)

Diagram P∗(λ) according to (4.32), (4.35) is represented in Fig. 4.16 by solidline. Replacing in (4.31) υ by 2λ − π/2 we get on the critical pressure (4.11)for the slope.

Pressure of Massif through Narrowing Channel

Similar to investigations of the previous subparagraph we can study thescheme in Fig. 4.17. We consider first the option 1 = h and the slip lines fieldconsisting of triangle AOB and sector OBC on each half. The parameters inthe triangle and on straight line OC are respectively

ψ = λ − π/4, σ3 = σm − τyi = −p∗;ψ = π/4, σ1 = σm + τyi = 0. (4.36)

Putting (4.36) into (4.27) we have

p∗ = 2τyi(1 + λ) (4.37)

and from static equation we finally receive

P∗ = 2lσyi(1 + λ)sinλ. (4.38)

Relation (4.38) is represented in Fig. 4.16 by broken line and we can see thatit is near to the solid curve which corresponds to the latter solution for b = 0.So, we can conclude that the simple results (4.37), (4.38) can be used for acase of l>h as well.


A

B C

O

h/2 h/2

|

P*

P*

Fig. 4.17. Pushing massif through channel

A

B

y

O

D

x

C

p*

p*

D’

Fig. 4.18. Pressure of punch and tension of plate with crack

At υ = π/2 in (4.31) and λ = π/2 in (4.37) we find the ultimate punchpressure (left part in Fig. 4.18) as

p∗ = σyi(1 + π/2). (4.39)

Tension of Plane with Crack

Relation (4.39) is valid for the problem of a crack in tension (right part inFig. 4.18). Here in square ODCD’

σx = σyiπ/2, τxy = 0, σy = σyi(1 + π/2)

and according to (2.72) we compute

σr = σyi(π/2 + sin2θ), σθ = σyi(π/2 + cos2 θ), τrθ = τyisin2θ. (4.40)

In the same manner we find in triangle AOB σy = τxy = 0, σx = σyi and


0

1

1

2

1

0

1 1

0

0 45 90 135

τrθ/σyi

Fig. 4.19. Diagrams of stresses

σr = σyi cos2 θ, σθ = σyi sin2 θ, τrθ = τyisin2θ. (4.41)

In sector OBD’ τrθ = τyi and stresses σr = σθ change as linear function of θ:

σr = σθ = σyi(0.5 + 3π/4 − θ). (4.42)

Diagrams σθ/σyi, σr/σyi, τrθ/σyi are represented in Fig. 4.19 by solid, brokenand interrupted by points lines 0. The same curves with index 1 refer to elasticsolution (3.107), at max τe = σyi/2. It is interesting to notice that these linesreflected relatively axis θ = π/2 describe the stress state near the punch edge.

It is also valid to note that in plastic state the potential function existsnear the crack ends as

0.5σyir2(π/2 + cos2 θ), 0.5σyir2(0.5 + 3π/4 − θ), 0.5r2σyisin2θ (4.43)

at θ ≤ π/4, π/4 ≤ θ ≤ 3π/4 and 3π/4 ≤ θ ≤ π respectively.

4.2.6 Ultimate State of Some Soil Structures

Conditions of Beginning of Plastic Shear

As we noticed above an earth is a very complex medium and its fractureis usually linked with shearing stresses. The strength condition is as a rulewritten in form τ < τ∗ - stable equilibrium, τ = τ∗ – ultimate state andτ > τ∗ - plastic flow where τ∗ is a characteristic of a material a value of whichdepends linearly on normal stress applied to the plane where τ acts. This isthe Coulomb’s law (here up to sub-chapter 4.3 according to /10/ compressivestresses are supposed positive with σ3 > σ1).

τ∗ = σ tan ϕ (4.44)

(inclined straight line in Fig. 1.22) for a quicksand and


τ∗ = σ tan ϕ + c (4.45)

(inclined broken line in the figure).- for a coherent soil. The latter equality isusually led to the form (4.44) (Fig. 4.20)

τ∗ = (σ + σc) tan ϕ (4.46)

where σc = c/ tan ϕ – coherent pressure which replaces an action of all cohesiveforces.

From (4.46) we have

tan ϕ = τ∗/(σ + σc). (4.47)

This condition may be written in another form. We draw through a point A(Fig. 4.21) at angle β to the horizon plane mn on which the components offull stress p - normal σβ and shearing τβ are acting. The first of them includesthe cohesion pressure. From geometrical consideration we find

τanθ = τβ/(σβ + σc). (4.48)

O

c

Fig. 4.20. Generalized Coulomb’s law

An

n

p

Fig. 4.21. Decomposition of full stress


Value of θ is usually called an angle of divergence which can not exceed angleof internal friction ϕ. That gives the condition of ultimate equilibrium as

θ = ϕ. (4.49)

Representations of Ultimate Equilibrium Condition

At an appreciation of materials’ strength the so-called Mohr’s circles are used.In the common representation of a tensor as a vector in a nine-dimensionalspace /10/ there are three such figures. At a plane stress state we have incoordinates σ, τ only one circumference (Fig. 1.22) along which a point moveswhen a plane turns in a material.

As was told in Chap. 2 the faces of a cube with absent shearing stresses arecalled main planes with normal stresses on them σ1 = σx, σ2 = σz, σ3 = σy.O. Mohr used his representation for a formulation of his hypothesis of strengthwhich in its linear option coincides with the Coulomb’s relation (4.44) and canbe interpreted as a tangent to the circumference in Fig. 1.22 under angle ϕ.

From expression (1.36) we have in main stresses the condition of theultimate state of quicksand as

sinϕ = (σ3 − σ1)/(σ3 + σ1). (4.50)

For coherent earth (4.50) can be generalized in form (broken line in Fig. 1.22)

sinϕ = (σ3 − σ1)/(σ1 + σ3 + 2ccotϕ). (4.51)

Relation (4.50) can be also represented in form

σ1/σ3 = tan2(π/4 ± ϕ/2). (4.52)

In the theory of interaction of structures with an earth sign minus correspondsto active pressure of soil and plus – to its resistance. In quicksand or coherentearth shearing displacements occur on planes under angles π/4 − ϕ/2 to thedirection of σ3.

In some cases it is useful to write (4.50), (4.51) in stresses σx, σy, τxy withthe help of (2.65) as follows

sin2 ϕ = ((σy − σx)2 + 4(τxy)2)/(σy + σx)2 (4.53)

for quicksand and

sin2 ϕ = ((σy − σx)2 + 4(τxy)2)/(σx + σy + 2c cot ϕ)2 (4.54)

for coherent soils.


DC

OK

p*p

*

h

B

A

l1

Fig. 4.22. Wedge pressed in soil

Wedge Pressed in Soil

We construct the field of slip lines as in Fig. 4.22 /25/ and we again supposethat OA is a straight line. From the figure we compute that it is inclined tohorizon AK by angle λ − υ as in Fig. 4.16. From geometrical considerationswe have 1 = a11 where

a = (1 − sin ϕ)(exp(−υ tan ϕ))/ cos ϕ

andh = 11(a cos λ − sin(λ − υ)). (4.55)

Putting (4.55) into the condition of constant volume similar to that for idealplastic material we find expression

h2 tan λ = (11)2 sin(λ − υ)(cos(λ − υ) + sin(λ − υ) tan λ)

which gives after transformations relation for tan λ:

(4a cos υ + sin 2υ) tan2 λ − 2(a2 + cos 2υ + 2a sin υ) tan λ − sin 2υ = 0. (4.56)

Now we find the ultimate load according to the field of slip lines in Fig. 4.22.From Fig. 1.22 we have for a cohesive soil

σ3σ1

= σm(1 ± sin ϕ) ± c cos ϕ. (4.57)

In triangle ABO σ1 = θ = 0 and from (4.57)

σm(1 − sin ϕ) = c cos ϕ

but from (4.28) for a cohesive soil σm = D − c/ tan ϕ and so

D = c/(1 − sin ϕ) tan ϕ. (4.58)

In the same manner for triangle OCD where σ3 = p∗, θ = υ we find from(4.57), (4.28)


p∗ = D(1 + sin ϕ)(exp 2υ tan ϕ) − c/ tan ϕ

and with consideration of D-value from (4.58) we receive finally

p∗ = c((1 + sin ϕ)e2υ tan ϕ/(1 − sin ϕ) − 1))/ tan ϕ. (4.59)

Lastly from static condition we derive

P∗ = 21c((1 + sin ϕ)e2υ tan ϕ/(1 − sin ϕ) − 1) sin λ/ tan ϕ. (4.60)

From diagrams P∗/21c = f(λ) at different ϕ in Fig. 4.23 we can see that P∗increases with a growth of ϕ and λ. It can be much bigger its value at idealplasticity (ϕ = 0, c = τyi – solid line in Fig. 4.16).

Some Important Particular Cases

At υ = π/2 − β we have from (4.59) the ultimate load for a slope (Figs. 3.5and 4.24) as follows

0

0

10

20

P* /2lc

Fig. 4.23. Dependence of P∗ on λ

p*

Fig. 4.24. Ultimate state of slope


p∗ = c((1 + sin ϕ)e(π−2β) tan ϕ/(1 − sin ϕ) − 1) cot ϕ (4.61)

and if υ = π/2 – well-known pu-value for a foundation (Fig. 3.12) – theso-called second ultimate load as

p∗ = (γeh + c cot ϕ)(1 + sin ϕ)eπ tan ϕ/(1 − sin ϕ) − c cot ϕ. (4.62)

4.2.7 Pressure of Soils on Retaining Walls

Active Pressure of Soil’s Self-Weight

A horizontal plane behind a vertical wall endures compression stress

σ3 = γez. (4.63)

Using equation of ultimate state (4.52) we find

σ1 = γez tan2(π/4 − ϕ/2). (4.64)

Diagram σ1(z) is given in Fig. 4.25 as triangle abd. The resultant of thispressure can be derived in form

Ra = 0.5γeH2 tan2(π/4 − ϕ/2). (4.65)

In the case of the earth’s passive resistance we must take in brackets ofexpressions (4.64), (4.65) sign plus.

When an uniformly distributed load q acts on a horizontal surface z = 0 weusually replace it by equivalent height h = q/γe (Fig. 4.26) and the resultantis

R = 0.5(σ1 + (σ1)′)H.

Since

σ1 = γe(H + h) tan2(π/4 − ϕ/2), (σ1)′ = γeh tan2(π/4 − ϕ/2) (4.66)

the resultant can be computed as

R = 0.5γeH(H + 2h) tan2(π/4 − ϕ/2). (4.67)

b

z

da

H

Ra

maxσ1

Fig. 4.25. Pressure of soil on vertical retaining wall


H

h

b

ad

q

Fig. 4.26. Additional external pressure

h

hC

H

Fig. 4.27. Consideration of coherence

Consideration of Coherence

If a soil has coherence its influence can be conditionally replaced by three-dimensional pressure of coherence σc = c/ tan ϕ (Fig. 1.22) and by equivalentlayer

h1 = σc/γe = c/γe tan ϕ. (4.68)

Taking this into account we can write

σ1 = γe(H + c/γe tan ϕ) tan2(π/4 − ϕ/2) − c/tan ϕ or

σ1 = γeHtan2(π/4 − ϕ/2) − 2c tan(π/4 − ϕ/2). (4.69)

According to Fig. 4.27 we can represent (4.69) as

σ1 = σ1ϕ − σ1c (4.70)

where σ1ϕ, σ1c are maximum lateral pressures in an absence of the coherenceand decrease of it due to coherent forces.


The whole pressure σ1 changes from tension in the top to compression inthe bottom and condition σ1 = 0 gives

hc = 2c/γe tan(π/4 − ϕ/2). (4.71)

The resultant of active pressure can be found as the area of shaded trianglewith base σ1 and height H − hc that is

Rc = 0.5σ1(H − hc). (4.72)

Putting in (4.72) σ1 according to (4.69) we compute

Rc = 0.5γeH2 tan2(π/4 − ϕ/2) − 2cH tan(π/4 − ϕ/2) + 2c2/γe.

Comparing this result to (4.65) we can conclude that the coherence maydiminish a resultant very strongly.

4.2.8 Stability of Footings

Besides the failures considered above a structure may loose its stability. Weconsider two types of such a phenomenon – plane and deep shears (Figs. 4.28and 4.29 respectively).

In the first case the loss of stability occurs by a movement parallel to hor-izontal surface, An appreciation of strength is usually made by a calculationof a factor of stability as

Ks = (fP + Ra)/Q (4.73)

where Q is a shearing force, f – coefficient of friction, P – weight of the struc-ture, Ra – resultant of active pressure computed by the relations (4.65), (4.67)and others of the previous paragraph.

In the second case the loss of stability takes place by a movement along acylindrical surface. The coefficient of stability can be calculated as a ratio ofsums of moments of resistance and shearing forces:

Ks =n∑

i=1

(Mi)res/n∑

i=1

(Mi)sh. (4.74)

P

QRa

Fig. 4.28. Plane shear


R

P

Ni

Pi

Ti

Fig. 4.29. Deep shear

To compute these sums we subdivide the soil massif by parts (blocks) andfind for each of them normal and tangent forces as

Ni = Pi cos αi,Ti = Pi sin αi. (4.75)

With consideration of relation (4.75) expression (4.74) can be represented inthe following way

Ks =

(n∑

i=1

Ni tan ϕ + cL

) /n∑

i=1

Ti (4.76)

where c is a specific coherence, L – a length of slip arc.

4.2.9 Elementary Tasks of Slope Stability

Soil has Only Internal Friction

We consider a slope inclined to the horizon under angle β (Fig. 4.30). ParticleM on its surface has weight P. We decompose it in normal N and tangent Tcomponents. Force T′ of friction resists to a movement of the particle. Fromthe equilibrium condition we have

P sin β = tan ϕPcos β

ortan β = tan ϕ. (4.77)

It means that ultimate angle β of a slope in quicksand is equal to its angle ofinternal friction ϕ.


T ′

T

P

N

M

Fig. 4.30. Equilibrium of particle on slope surface

T

T ′

P

DN

z

Fig. 4.31. Influence of filtration pressure

Influence of Filtration Pressure

The angle of internal friction depends on hydrodynamic pressure D of a waterin a condition of its filtration. In this situation shearing forces are (Fig. 4.31)

T = P sin β,D = γwni sin β (4.78)

where γw is a specific weight of the water, n – porosity, i sin β–a hydraulicgradient. Resistance force is

T′ = P′ cos β tan ϕ. (4.79)

Here P′ = (γe)′i and (γe)

′ is a specific weight of soil suspended in the water.With consideration of (4.78), (4.79) the stability factor is

Ks = T′/(T + D) = (γe)′ tan ϕ/((γe)

′ + γwn) tan β. (4.80)

Coherent Soil

Now we consider a vertical slope of coherent earth when slip surface is aplane (Fig. 4.32).


a

h

T

cb

P

N

Fig. 4.32. Vertical slope of coherent soil

The acting force is self-weight P of sliding prism abc as

P = 0.5γeh2 cot β (4.81)

from whichT = 0.5γeh

2 cos β (4.82)

The force of resistance isT′ = hc/ sin β.

In ultimate state T = T′ and with consideration of (4.82) we derive

0.5h2γe cos β = hc/ sin β (4.83)

from whichc = (γeh/4) sin 2β. (4.84)

According to condition of ultimate equilibrium β = π/4−ϕ/2 and at ϕ = 0the slip plane makes with the horizon angle π/4. Taking this into account wefind ultimate height of the vertical slope as

h = 4c/γe.

4.2.10 Some Methods of Appreciation of Slopes Stability

Rigorous Solutions of Ultimate Equilibrium Theory

A rigorous solution of slope stability takes into account both the angle ofinternal friction and the coherence. Two main cases should be considered.

1. Maximum vertical pressure is given by relation (4.61) which correspondsto plane slope. Here some special tables are also used at different β, ϕ and


dimensionless ultimate pressure σo can be found. Then the whole ultimatepressure with consideration of coherence can be written as follows

pu = σo + c cot ϕ,

2. A slope in its ultimate state can support on its horizontal surface uni-formly distributed load with intensity

p∗ = 2c cos ϕ/(1 − sin ϕ).

This value can be considered as an action of an equivalent soil’s layer with aheight

h = 2c cos ϕ/γe(1 − sin ϕ).

When c and ϕ both are not equal to zero a construction of most equallystable slope may be fulfilled by the Sokolovski’s method of dimensionlesscoordinates

x = Xc/γs, y = Yc/γs (4.85)

beginning from the top of the slope.

Method of Circular Cylindrical Surfaces

The method consists in a determination of a stability coefficient of naturalslope for the most dangerous slip surfaces. In practice they are taken circularcylindrical and by a selection of the centre of the most dangerous one (forwhich Ks has minimum) is found.

Let the centre be in a point O (Fig. 4.33). We draw from it through thelower point an arc of slip and construct the equilibrium equation for massifabd. For this purpose we divide it by vertical cross-sections in n parts and usecondition ΣM = 0 as

a

c

db

h

R

O

TiNiPi

Fig. 4.33. Circular cylindrical surface


4.5h

2h

h

d

b

a

O2O1

Oc0

c1

c2

cmax

Fig. 4.34. Search for the most dangerous sliding surface

n∑i=1

TiR −n∑

i=1

NiR tan ϕ − cLR = 0. (4.86)

Excluding from (4.86) R we have

Ks =

(n∑

i=1

Ni tan ϕ + cL

) /n∑

i=1

Ti. (4.87)

To receive the most dangerous surface we behave in the following way(Fig. 4.34). We begin with case ϕ = 0 and find point O using angles β1,β2 from Appendix D. Then we put points O1, O2, . . . at equal distances andcompute for each of them c-values according to (4.86) for consequent slidingsurface. cmax corresponds to the most dangerous slope.

A simplification of this method was made by Prof. M. Goldstein accordingto whom

Ks = Atan ϕ + Bc/γeh (4.88)

where coefficients A, B must be taken from Appendix E. It is not difficult tofind h (Fig. 4.35) as

h = cB/γe(Ks − Atan ϕ) (4.89)

Method of Equally Stable Slopes by Approach of Professor Maslov

The method is based on the supposition that at the same pressure angle ϕof resistance to shear in laboratory tests is linked with angle of repose ψ innatural conditions as

tan ψ = tan ϕ + c/γeH. (4.90)

To construct a profile of a stable slope we divide it on a row of layers(Fig. 4.36) and compute for each of them the pressure of the soil on a lower


h

l

C1C2C3358

1:m

0.4m0.3h

0.3h0.3h

Fig. 4.35. Method of M. Goldstein

a

H1

H2

H3

H4

b

c

d

e

Fig. 4.36. Profile of equally stable slope

plane and angle of shear by (4.90) with consideration of stability coefficientas follows

tan ψ = (tan ϕ + c/p)/Ks. (4.91)

The profile of such a slope with computed values of ψ beginning from thelower layer is given in Fig. 4.36.

Method of Leaned Slopes

This method is used for an appreciation of landslide stability at fixed slipslopes and stability coefficient is computed according to (4.87). For a choiceof a place of prop structure a pressure of a landslide must be found by thisway. The massif is divided by parts (blocks) and for each of them the slipsurface is a plane. According to the equilibrium condition for each of them(Fig. 4.33)

Ri + Ni tan ϕ + cLi − Ti = 0 (4.92)

4.3 Axisymmetric Problem 117

we have

R1 = P1 sin α1 − P1 cos α1 tan ϕ1 − c1L1,

R2 = P2 sin α2 − P2 cos α2 tan ϕ2 − c2L2 + R1 cos(α1 − α2)

or

Ri = Pi sin αi − Pi cos αi tan ϕi − ciLi + Ri−1 cos(αi−1 − αi) (4.93)

where Ri−1 is the projection of landslide pressure of preceding part on thedirection of slip of the block in the consideration and in the point with Rmin

corresponds to the place of the prop structure.

4.3 Axisymmetric Problem

4.3.1 Elastic-Plastic and Ultimate States of Thick-WalledElements Under Internal and External Pressure

Sphere

We begin with a sphere (Fig. 3.23) and computing difference of stresses σθ −σρ from (3.114) and equalling it to σyi we find the difference of pressureswhich corresponds to the beginning of plastic deformation at ρ = a (q > p iseverywhere in this paragraph) at β = b/a as follows

(q − p)yi = 2σyi(1 − β−3)/3. (4.94)

When q−ρ > (q − p)yi we have two zones – an elastic in c ≤ ρ ≤ b where

σρ = C1 + C2/ρ3, σθ = C1 − C2/2ρ3

and a plastic one at a ≤ ρ ≤ c. In the latter we determine from (2.80), bound-ary condition σρ(a) = −q and yielding demand σθ − σρ = σyi – expressions

σρ = −q + 2σyi ln(ρ/a), σθ = −q + σyi(1 + 2 ln(ρ/a)). (4.95)

Constants C1, C2 can be excluded according to conditions σρ(b) = −p andσθ − σρ = σyi at ρ = a. As a result we have in the elastic zone

σρ = −p + 2c3σyi(1− b3/ρ3)/3b3, σθ = −p + 2c3σyi(1 + b3/2ρ3)/3b3. (4.96)

From the compatibility law for stresses at ρ = c we find the dependence ofp − q on c and its ultimate value at c = b as follows

q − p = 2σyi(1 − c3/b3 + 3 ln(c/a))/3, (q − p)u = 2σyi ln β. (4.97)


Cylinder

It can be considered in the same manner. From (3.104) we find the differenceof pressures at which the first plastic strains appear as

(q − p)yi = σyi(1 − β−2). (4.98)

At (q − p)yi ≤ (q − p) we have in the elastic zone c ≤ r ≤ b

σr = −p + c2σyi(1 − b2/r2)/2b2, σθ = −p + c2σyi(1 + b2/r2)/2b2.

The dependence of q − p on c and its ultimate value at c = b are

q − p = 0.5σyi(1 − c2/b2 + 2 ln(c/a)), (q − p)u = σyi ln β. (4.99)

In plastic zone the expressions for stresses are similar to (4.95) and they canbe written as follows

σr = −q + σyi ln(r/a), σθ = −q + σyi(1 + ln(r/a)).

Comparing (4.94), (4.97) to (4.98), (4.99) respectively we can concludethat a sphere demands the smaller difference of pressures for the beginning ofyielding and twice of it at ultimate state than a cylinder.

Cone

As in the case of the cylinder (Fig. 3.24) the first plastic strains appearaccording to (3.118) at

(q − p)yi = 0.5Aσyi sin2 ψ/ cos ψ

where A is given by (3.120). At q − p > (q − p)yi we have in elastic and plasticzones, respectively

σθσχ

= −p + 0.5σyi sin 2υ(cos λ/ sin2 λ ± cos χ/ sin2 χ

+ ln(tan(λ/2)/ tan(χ/2))),σχ = −q + σyi ln(sin χ/ sin ψ), σθ = −q + σyi(1 + ln(sin χ/ sin ψ)).

The dependence of q − p on angle υ at the border between elastic andplastic zones and the ultimate state are described by relations

q − p = 0.5σyi(1 + 2 ln(sin υ/ sin ψ − sin2 υ(cos υ/ sin2 υ + ln(tan(λ/2)/ tan(υ/2))),

(q − p)u = σyi ln(sin λ/ sin ψ). (4.100)


Let us now consider the yielding of the cone with initial angles ψo, λo when itis in the ultimate state. From (2.70) we derive the constant volume equationfor velocity V = uχ/ρ in following form

dV/dχ + V cot χ = 0

with obvious solutionV = sin ψ/ sin χ.

But according to definition V = dχ/dψ that gives the integral which can bealso found from the condition of constant volume of differences of sphericalsectors

cos χ − cos χo = cos ψ − cos ψo (4.101)

and instead of (4.100) we can write

(q − p)u = 0.5σyi ln(1 − (cos ψ − cos ψo + cos λo)2/ sin2 ψ).

Sokolovski /18/ investigated also the case of π/2 ≤ λ when two plasticzones (AOB and COD in Fig. 4.37) appear. This problem can be consideredsimilarly to the previous one.

Particularly the ultimate state takes place at υ = π/2 and for it

(p − q)u = σyi ln(sinψ/ sin λ).

4.3.2 Compression of Cylinder by Rough Plates

L. Kachanov found upper load compressing a cylinder of height 2h andradius r (Fig. 4.38). He proposed displacements in form

ur = U = Cr(1 − βz/h),uz = V = Voz/h

A

B

C

Dp

q

O

Fig. 4.37. Cone with big angle at apex


a aP

P

r

zh

h

Fig. 4.38. Compression of cylinder

01

2

3

4

4 8 12 a/h

Fig. 4.39. Dependence of ultimate pressure on a/h

where β is a barrel factor, Vo – velocity of the plates and constant C can befound from the constant volume demand (see expressions (2.76)) as follows

dU/dr + U/r + dV/dz = 0.

Then L. Kachanov uses the equality of power of external and internalforces as

P∗Vo = 2πτyi

⎛⎝

η∫

0

1∫

0

εeρdρdξ +

1∫

0

U(η)ρdρ

⎞⎠ . (4.102)

Here η = h/a, ρ = r/a, ξ = z/a and effective strain εe is given by relation(2.25). Computing εr, εz, εθ, γrz according to (2.76), putting it into (4.102) weget on a complex expression. Parameter β should be found from the conditionof minimum p∗ where p = P/πa2. The dependence of p∗ on a/h is shown bysolid curve in Fig. 4.39. The broken line in the figure corresponds to elementarysolution β = 0 in form


p∗/σyi = 1 + 1/3√

3η.

It is easy to see that the simple solution gives results near to the rigorousones. In a similar manner the problems of stress state finding can be fulfilledfor a neck in a bar at tension.

4.3.3 Flow of Material Within Cone

Common Case

Similarly to Sect. 4.2.2 we consider a flow within a cone (Fig. 3.6 where coor-dinates r, θ must be replaced by ρ, χ respectively). Above that we supposehere τ ≡ τχθ and that strains εχ = εθ = −ερ/2 depend only on χ. Removingfrom (2.78) difference σρ − σχ according to (2.65) and condition τe = τyi weget on the first integral as

dτ/dχ = 2nτyi − τ cot χ − 4√

(τyi)2 − τ2

where n is a constant. Putting here representation (4.13) at r = ρ, θ = χ weget equation

dψ/dχ = n/ cos 2ψ − 2 − 0.5 cot χ tan 2ψ (4.103)

that cannot be solved rigorously. Sokolovski /18/ gave diagrams λ(n) andψ(χ) for λ ≤ 40.

These curves for 0 ≤ λ ≤ π are represented by solid lines in Figs. 4.40and 4.41.

Now from the second equation (2.77) and representation (4.13) at τe = τyi

we deriveσχ = F(ρ) − 3τyi

∫sin 2ψdχ

where F is a function of ρ. Computing from (4.13) stress σρ and putting it intothe first (2.77) we find with consideration of (4.103) F(ρ) and hence stressesdepending on constant C.

20

0.8

1.6

2.4

2.5 3 3.5 n

Fig. 4.40. Diagram λ(n)


0

0.2

0.4

0.630 75 115

149

179

0.8 1.6 2.4

Fig. 4.41. Diagrams ψ(χ) at different λ

σχ = C − τyi(2n ln ρ + 3∫

sin 2ψdχ), σρ = σχ + 2τyi cos 2ψ. (4.104)

Case of Big n

In /18/ Sokolovski proposed some simplifications as in Sect. 4.2.2 and gotsolution for small λ as

λ = 1/2n, τ = τyiχ/λ, σρ − σχ = (2τyi/λ) ln(a/ρ),

σρ = 2τyi(λ−1 ln(a/ρ) + 2√

1 − (χ/λ)2),V = Vo(1 + 2√

λ2 − χ2)

Diagram λ(n) according to the first of these expressions is drawn in Fig. 4.40by broken line and we can see that ii is valid only for very small λ.

Approximate Approach

If we neglect in (4.103) the last member we can integrate the equationrigorously as follows

χ = 0.5(n(n2 − 4)−0.5 tan−1(√

(n + 2)/(n − 2) tan ψ) − ψ).

(interrupted by points lines in Fig. 4.41). At ψ = π/4 this relation givesexpression for λ(n) – broken-solid curve in Fig. 4.40). From (4.13), (4.104)and integral static equation

λ∫

0

σρ(a, χ) sin 2χdχ = 0 (4.105)


0

5

10

15

J

22.5 67.5 λo45

Fig. 4.42. Diagram J(λ)

we have approximately

σχσρ

= τyi(2n ln(a/ρ) − J/8 sin2 λ − 0.375n ln(n − 2 cos 2ψ)−0.75+1.25x cos 2ψ)

where (Fig. 4.42)

J =

λ∫

0

(10 cos 2ψ − 3n ln(n − 2 cos 2ψ)) sin 2χdχ.

If we suppose σρ(a, λ) = σχ(a, λ) = −q∗ then we find

q∗ = τyi(3n ln n + J/ sin2 λ)/8 (4.106)

and this expression will be used later in Chap. 5.

4.3.4 Penetration of Rigid Cone and Load-Bearing Capacityof Circular Pile

These problems can be solved in the same manner as in Sect. 4.2.4. Thebasic relations of the previous subparagraph are valid here but instead of(4.105) we must use similar to (4.20) (Fig. 4.9 with consequent replacementof coordinates) integral static laws for axisymmetric problem

p∗b2 = 2a2

λ∫

0

(σρ(a, χ) cos χ + τ(a, χ) sin χ) sin χdχ,

P∗/π = p ∗ b2 + 2 sin λa+1∫

a

(σχ(ρ, λ) sin λ + τ(ρ, λ) cos λ)ρdρ.

(4.107)


As a result we find

P∗/π = (a + 1)2 sin2 λ + τyi(1(1 + 2a)(J1/4 + cos λ sin λ

+ n(1 − (3/8) ln n) sin2 λ) − 2n(1 + a)2 ln(1/a + 1) sin2 λ). (4.108)

Here

J1 = −λ∫

0

(10 cos 2ψ − 3n ln(n − 2 cos 2ψ) cos χ + 8 sin 2ψ sin χ) sin χdχ.

For λ near to π and a → ∞ (the option of a circular pile) we compute

P∗/π = p∗b2 + 2τyibl(1 + J1/4 sin λ). (4.109)

The calculations at n = 2.045, λ = 179 give J1 ≈ 6.5 and hence ratioJ1/4 sin λ is big (due to an approximative character of the above theory)Neglecting this member we have a simple result in a safety side as following

P∗ = πb(p∗b + 2τyi1).

4.4 Intermediary Conclusion

An importance of the results according to the scheme of the perfect plasticbody is difficult to overestimate. They give ultimate loads and therefore –the moment of structures destruction. We can also notice that a procedureof computations is much simpler that by the Theory of Elasticity methods.However the solutions of this chapter can be used first of all for the materialswith big yielding part of stress-strain diagram or small hardening (for soils– with distinctive angle of internal friction and cohesiveness). Above thatthe process of deformation and fracture between elastic and plastic stages isunknown. This gap can be removed in computations according to equationsof the hardening body which is considered in the following two chapters whereunsteady non-linear creep and damage are also taken into account.

5

Ultimate State of Structures at SmallNon-Linear Strains

5.1 Fracture Near Edges of Cracks and Punchat Anti-Plane Deformation


Equilibrium equation (2.56) is satisfied if we take

τr = −∂W/r∂θ, τθ = ∂W/∂r

where W is a potential function that can be written in form W = Krsf(θ)which gives relations for stresses

τr = −Krs−1f ′(θ), τθ = Ksrs−1f(θ),

τe = Krs−1√

(sf)2 + (f ′)2.(5.1)

Since /21/ τiγi is proportional to r−1 we find s = m/(m + 1) and according to(2.57), (5.1) – derive following values

uz = Ω(t)(m + 1)Kmr1/(m+1)((sf)2 + (f ′)2)(m−1)/2f ′,

τθ = Kmr−1/(m+1)f/(m + 1).(5.2)

From compatibility expression for strains (2.58) we have differential equa-tion for f(θ) as /16/

f ′′+(1+m(m−1)(((m+1)f ′)2+f2)/(((m+1)f ′)2+mf2))f/(m+1)2 = 0. (5.3)

At m = 1 we have from (5.3) results (3.10) and (3.19).In order to get a rigorous solution we put in (5.3)

f = ez, f ′ = ezU, f ′′ = ez(U2 + U′)

that gives (C is a constant)

126 5 Ultimate State of Structures at Small Non-Linear Strains

U = m(√

(1 + m)2 + 4 tan2(C − θ) − (1 + m))/2(1 + m) tan(C − θ). (5.4)

On the other hand we have ln(f/D) = Udθ (D is also a constant) and withconsideration of (5.4) as well as supposing tan(C − θ) =

√x we receive

4dln(f/D) = (1 −√

(m + 1)2 + 4mx)dx/x(1 + x). (5.5)

and after integration we find

f = D((√− m + 1)/(

√+ m − 1))(m−1)/4(m+1)

((√

+ m + 1)/(√− m − 1))1/4

√sin(C − θ) (5.6)

where√ =

√(1 + m)2 + 4 tan2(C − θ).

5.1.2 Case of Crack Propagation

For this task from boundary conditions f(0) = 1, f(π) = 0 (Fig. 3.4 and rela-tions (5.1)) we compute C=0 and the value of another constant

D = mm/2/(1+m)/(1 + m)0.5. (5.7)

From Fig. 5.1 where according to expressions (5.1) and (5.6) diagrams τr(θ),τθ(θ) are constructed for m = 1, 3, 15 by solid, broken and interrupted bypoints lines we can see that with an increase of m the role of one componentof τe in the certain part of the plane is growing.

In order to find the value of K we compute integral J = dΠ/dl where

dΠ =

dl∫

0

τθ(r, 0)uz(dl − r, π)dr

0

0.25

0.5

0.75

τθ/τyi

τθτr/τyi

τr

45 90 135 θ°

Fig. 5.1. Diagram of shearing stresses near crack end

5.1 Fracture Near Edges of Cracks and Punch at Anti-Plane Deformation 127

is a free energy of crack propagation. Taking into an account uz and τθ from(5.2) we calculate

J = Ω(t)Km+1m(f ′(π))msignf ′(π)

1∫

0

(ξ/(1 − ξ))1/(1+m)dξ.

According to /21/ a value of J in the similar problem for tension does notdepend on the properties of a material. Assuming this supposition also for ourtask we can equal J to its meaning at m = 1 that gives

Km+1 = −τo2πlsignf ′(π)/2GΩ(t)/f ′(π)/mI(m)

where

I(m) = (m + 1)Γ((2 + m)/(1 + m))Γ((2m + 1)/(m + 1)).

Here Γ() is gamma-function and f ′(π) must be taken from solution (5.6).Computing τe by the third expression (5.1) we receive the following equationfor r(θ) at τe = constant

2rGΩ(τe)1+mr/(τo)2l = π(√

(sf)2 + f ′2)m+1/I(m)/f ′(π)/m. (5.8)

From Fig. 5.2 where condition (5.8) is represented by solid, broken and inter-rupted by points lines for m = 1, 3, 15 respectively (see Appendix F) we cansee that with the growth of m plastic zone increases and moves out of thecrack. It confirms the solution of Sect. 4.1.2 for a perfect plastic body.

5.1.3 Plastic Zones near Punch Edges

In this case we have boundary conditions f(0) = 0, f ′(π) = 1 and from (5.6) wefind C = π as well as the same D given by (5.7). It means that in the previous

−0.2 0 0.3 0.5

rθ

Fig. 5.2. Plastic zones for non-linear material


solution angle θ must be replaced by value π − θ and diagrams in Figs. 5.1,5.2 should be reflected relatively to axis θ = π/2. So, with a growth of m theinelastic zone moves under the punch and this corrects the position of theplastic district for the same problem in Sect. 4.2.1.

As a conclusion we must add that the results in this paragraph are validfor small plastic zones in which the asymptotic relations (3.10), (3.19) can beused. Rice in /21/ gave an approximate solution also for big inelastic districts.


5.2.1 Generalization of Flamant’s Problem

Common Solution

As in Sect. 3.2.3 we suppose σθ = τrθ = 0 (Fig. 3.11) that allows to find fromthe first static equation (2.67)

σr = f(θ)/r (5.9)

and according to rheological law (2.66) at σe = σx = σr we have

εr = −εθ = 0.75Ω(t)r−mg(θ) (5.10)

wherein g(θ) = fm. From (2.66) at α = 0 and condition γrθ = 0 compatibilityexpression (2.71) becomes

g′′ + β2g = 0. (5.11)

Here β =√

m(2 − m). The solution of (5.11) depends on the value of m /18/.

Particular Cases

If m = 2 theng = Cθ + D

and due to symmetry condition C = 0. The constant D can be found fromintegral static laws that for λ = π/2 (see Fig. 3.11) are

P cos αo = −π/2∫

−π/2

rσr cos θdθ, P sin αo = −π/2∫

−π/2

rσr sin θdθ. (5.12)

So, in this particular case D = (P/2)2 and stress

σr = P/2r

does not depend on angle θ (at αo = 0 in Fig. 3.11).


In a similar way cases m > 2,m < 2 can be studied. Taking into an accountthe symmetry condition we receive respectively

σr = D1(cosh βθ)µ/r, σr = D2(cos βθ)µ/r. (5.13)

Constants D1,D2 can be found from expressions (5.12). At m = 1 and αo = 0we receive the Flamant’s result

For practical purposes it is interesting to establish the dependence of stressσy on angle θ. From (2.72), (5.12), (5.13) we have for m = 1, 2, 4 respectively

σy(1) = −(2P/πy) cos4 θ, σy(2) = −(P/2y) cos3 θ,

σy(4) = −0.4(P/y)(cosh 2√

2θ)1/4 cos3 θ.(5.14)

Corresponding diagrams /σy(θ)/ are constructed in Fig. 5.3 by solid, brokenand interrupted by points curves. We can see that with an increase of m thestress distribution is more even.

Comparison of Results

In order to appreciate the results we compare for the case m = 1 the distrib-ution of stresses σy along vertical axis under the concentrated load as well ascentres of the punch and uniformly placed load where we have according tothe first relation (5.14) (solid line in Fig. 5.4), and expressions (3.95) (brokencurve in the figure), (3.52) (interrupted by points line) respectively

σy/p = −4l/πy, σy/p = −2(1 + 2(y/l)2)/π(1 + (y/l)2)3/2,

σy/p = −2(tan−1(l/y) + (y/l)/(1 + (y/l)2))/π.(5.15)

Here p = P/2l and we can see from Fig. 5.4 that at y > 3l the curves practicallycoincide. Since with the growth of non-linearity the stress distribution becomesmore uniform we can expect that solutions (5.13) can replace other forms ofpressure on the foundation at least at y > 3l.

0

0.2

0.4

0.6

σyy/P

π/6 π/3 θ

Fig. 5.3. Distribution of stresses at different m


0 0.4 0.8 σy/p

2

4

y/l

Fig. 5.4. Distribution of stresses under different loads

0.64 0.5 0θ

0

Q

0.5

0.96

Fig. 5.5. Distribution of stresses due to horizontal force

Case of Horizontal Force

The results of the previous subparagraph can be used here for the caseλ = αo = π/2 if we compute angle θ from horizontal direction when we havefor m = 1, 2, 4 respectively (solid, broken and interrupted by points lines inFig. 5.5)

σr = −2(Q/πr) sin θ, σr = −Q/2r, σr = −0.4Q(cosh 2√

2θ)1/4/r. (5.16)

In order to give to the results just received a practical meaning we comparefor the case m = 1 the distribution of σr along axis x according to the firstrelation (5.16) and expressions (3.101), (3.109) as follows


x/l 5 3 1

0.4

0.8

σr/q

Fig. 5.6. Distribution of stresses at different loadings

σr = −4ql/πx, σr = −4q((x/l)2 − 1)−1/2/π.

where q = Q/2l, and from Fig. 5.6 in which the corresponding diagrams aregiven by solid and broken lines we can see that the curves are near to eachother and practically coincide at x/l > 3. So, we can use results (5.16) in anon-linear state at least out of this district.

5.2.2 Slope Under One-Sided Load

General Relations

For the purpose of this paragraph we rewrite (2.66) at α = 0, γrθ ≡ γ infollowing form (see also (2.30))

σr − σθ = 4ω(t)(γm)µ−1εr, τ = ω(t)(γm)µ−1γ (5.17)

where γm is a maximum shearing strain

γm =√

(εr − εθ)2 + γ2

linked with τe by the law similar to (2.30) as

τe = ω(t)(γm)µ. (5.18)

Putting (5.17) and similar to (4.13) representations for strains

εr = 0.5γm cos 2ψ, γ = γm sin 2ψ (5.19)

into (2.68) and the third equation (3.21) we get on the system

d((γm)µ sin 2ψ)/dθ + 2(γm)µ cos 2ψ = 0, (5.20)

d(γm cos 2ψ)/dθ = 2γm sin 2ψ + C1 (5.21)

where C1 is a constant. At µ = 1 we have from (5.20), (5.21) the solution ofSect. 3.2.1.


Fulfilling the operations in (5.20), (5.21) and excluding γm we receive thesecond order differential equation which is not detailed in /18/. Replacingin it

Θ = −dθ/dψ (5.22)

we find the first order differential equation

(tan 2ψ)dΘ/Θdψ = 2(1 − Θ)((Θ − 1)/µ + 1 − 2/Ψ) (5.23)

in whichΨ = 1 − (1 − µ) sin2 2ψ. (5.24)

Sokolovski /18/ gave curves ψ(θ), τe(θ) and max τe(λ) for µ = 1/3,λ < π/4. The latter is shown by broken line in Fig. 5.7 (we calculated ittill λ = π/3). Solid curve refers to expression (3.25) (for a linear material atθ = 0).

Here we integrate (5.23) by the finite differences method at boundarycondition Θ(0) = 1. Then we integrate (5.22) at border demand θ(0) = λ.Another similar condition θ(π/4) = 0 allows to choose ratio (1 − Θ)/ tan 2ψin point Θ(0) = 1. The calculations were made by a computer.

Results of Computation

Firstly we consider case µ = 1 when we have (solid curves in Fig. 5.8)

tan 2ψ = (cos 2θ − cos 2λ)/ sin 2θ. (5.25)

At λ = π/4 and λ = π/2 we can receive from (5.25) straight lines

ψ = −θ + π/4,ψ = −θ/2 + π/4

respectively. Differentiating (5.25) we find

300.2

0.4

0.6

max τe/p

45 60 75 λ°

Fig. 5.7. Dependence of max τe on λ


0

20

ψ°

30 60

π/2π/3

π/4

θ°

Fig. 5.8. Dependence of ψ on θ at different λ

Θ = 1 − (sin 2ψ)(tan2 2λ + sin2 2ψ)−1/2

or after transformations as a function of θ

Θ = (1 + cos2 2λ − cos 2λ cos 2θ)/(1 − cos 2λ cos 2θ). (5.26)

The approximate calculations reveal good agreement with (5.25), (5.26). Itallows to use the finite differences method for another µ. The curves forµ = 1/3 and µ = 2/3 practically coincide with solid lines in Fig. 5.8.

When function ψ(θ) is known a value of τe can be found from equationsfollowing of (3.21), (4.13) and boundary conditions for σθ (see Sect. 3.2.1) as

dτe/τedθ + 2(dψ/dθ + 1) cot 2ψ = 0,

p = 4λ∫0

τe sin 2ψdθ.

Combining these expressions we find for max τe = τe(0) at λ ≥ π/4

p = 4max τe

λ∫

0

sin 2ψ exp(−2

θ∫

0

(1 + dψ/dθ) cot 2ψdθ)dθ. (5.27)

Computations for µ = 2/3, 1/2 and 1/3 show that diagrams max τe(λ) arenear the solid line in Fig. 5.7. It can be explained by the absence of µ in (5.27)and the vicinity of curves in Fig. 5.8 at different µ. It allows to use the solidlines in the latter figure for practical purposes.

In order to find the ultimate state we rewrite (5.18) with consideration of(1.45), (2.30) on axis θ = 0 where εθ = εr = 0, γ = 2ε1 ≡ 2ε > 0

Ω(t)(2max τe)m = εe−α. (5.28)

Using the criterion dε/dt → ∞ we receive the values at critical state as

ε∗ = 1/α,Ω(t∗) = (2max τeeα)−m. (5.29)


Simple Solution

In order to find engineering relations we rewrite (5.17) in form of (2.66) atα = 0 as follows /25//

εr = Ω(t)(τe)m−1(σr − σθ)/4, γ = Ω(t)(τe)m−1τ (5.30)

and put them into the third (3.21) that gives

(τe)m−1((m − 1)(τe)−2τ′2 + 4)(τ′′ + 4τ) = C2. (5.31)

where C2 is a function of t. Here and further the dependence on time is hinted.According to the symmetry condition τ′(0) = 0 and taking this assumptionfor the whole wedge we receive from (5.31)

(τe)m−1(τ′′ + 4τ) = C2/4

which gives at m = 1 the solution of Sect. 3.2.1.To exclude C2 from (5.31) we differentiate it as follows

(m − 1)((m − 3)(τ′′ + 4τ)τ′2 + 4(τe)2(3τ′′ + 4τ))τ′(τ′′ + 4τ)+ 4(τe)2((m − 1)τ′2 + 4(τe)2)(τ′′′ + 4τ′) = 0.

(5.32)

Here we again suppose τ′ = 0 in the whole wedge that gives from (5.32) τ′′′ = 0and with consideration of (3.21) as well as the same boundary conditions asin Sect. 3.2.2 for σθ, τ at ±λ we compute

τ = 3p(λ2 − θ2)/8λ3, σr − σθ = 3pθ/4λ3,

σθσr

= 3p(θ3/3 + θx−λ2

1−λ2)/4λ3 − p/2, τe = 3p√

θ2 + (θ2 − λ2)2/8λ3. (5.33)

From Fig. 5.7 we can see that interrupted by points curve corresponding to(5.33) at θ = 0 may be taken as the first approach.

5.2.3 Wedge Pressed by Inclined Rigid Plates

Engineering Relations for Particular Case

We considered the problem of a pressed wedge in Sects. 3.2.2 and 4.2.2 forelastic and plastic media. Here we study the task for a hardening at creepmaterial and begin with the case of parallel moving plates (Fig. 3.6) at negli-gible compulsory flow /23/. From (3.28) we have at uθ = −V(θ)

ur = V′, εr = εθ = 0, γ = (V′′ + V)/r (5.34)

and using (5.17), (2.67) we find

τ = ω(t)r−µf(θ), σr = σθ = F(r) − ω(t)r−µ(2 − µ)∫

f(θ)dθ (5.35)


where F is a function of r and

f(θ) = (V′′ + V)µ.

Putting (5.35) into (2.68) we get on equality

ω(t)(µ(2 − µ)∫

f(θ)dθ + f ′(θ)) = −r1+µdF/dr

which is true if its both parts are equal to the same function of t, say n(t).This gives two expressions

F = A − mr−µn, f ′′ + µ(2 − µ)f = 0.

Taking into account the symmetry condition we write the solution of the latterequation as following

f(θ) = C sin βθ.

Here β =√

µ(2 − µ) and n = 0. So, from expressions (5.35) and integral staticlaws (3.32) we derive

τ = Cω(t)r−µ sin βθ, σr = σθ = −Cω(t)a−µ(L − (a/r)−µ cos βθ)mβ (5.36)

where

Cω(t) = qaµ/mβ(L − H cos βλ),L = ((sin(β − 1)λ)/(β − 1) + (sin(β + 1)λ)/(β + 1))/2 sin λ,H = ((l/a + 1)1−µ − 1)a/l(1 − µ).

At µ = 1 we have from (5.36) solution (3.40). The dependence of max τe

on λ is shown by solid line 0.5 in Fig. 3.9 for µ = 0.5. Diagrams σθ(r) also forµ = 0.5 are given by solid curves 0.5 in Figs. 3.7, 3.8 for λ = π/4 (a model ofa retaining wall) and λ = π/2 (a flow of a material between two foundations).At λ → 0 we get the solution near to that in /26/.

Now we find the critical state according to the scheme of Sect. 5.2.2. Inthe current task we have also εr = εθ = 0, ε ≡ ε1 = γ/2 > 0 and accordingto the criterion of infinite elongation rate dε/dt → ∞ at dangerous pointsa = r, θ = λ we compute with a consideration of (5.36), (5.28)

ε∗ = 1/α,Ω(t∗) = (αeq sin βλ/mβ(L − Hcos βλ)−m.

Flow of Material between Immovable Plates

Now we consider the case when only the compulsory flow takes place (a modelof a volcano row) and here we have from (3.28)

ur = U(θ)/r, εθ = −εr = U(θ)/r2, γ = dU/r2dθ (5.37)

and according to (5.18) we find


γm = g(θ)/r2 (5.38)

whereg =

√U′2 + 4U2. (5.39)

Using the representation similar to (5.19)

εr = (g/2r2) cos 2ψ, γ = (g/r2) sin 2ψ (5.40)

we have from (5.37), (5.40)

ln(/U/ : D) = −2

θ∫

0

tan 2ψdθ, g = −2U/ cos 2ψ. (5.41)

Here D is a constant that will be found later. We must also notice that thesolution satisfies stick condition U(λ) = 0.

Putting (5.40) into compatibility law following from (5.37) as ∂εθ/∂θ = γwe receive equation

(g cos 2ψ)′ + 2g sin 2ψ = 0 (5.42)

which also gives the boundary condition

dψ/dθ = 1 (5.43)

at θ = λ. Above that we find from (5.42) expression for g(θ) as

ln(g/D) = 2

θ∫

0

(dψ/dθ − 1) tan 2ψdθ. (5.44)

Now from (5.17) and (5.40) we derive expressions

σr − σθ = 2ω(t)r−2µgµ cos 2ψ, τ = ω(t)r−2µgµ sin 2ψ (5.45)

which together with (2.68) give

(gµ sin 2ψ)′′ + 2(1 − 2µ)(gµ cos 2ψ)′ + 4µ(1 − µ)gµ sin 2ψ = 0. (5.46)

From (5.46), (5.42) we find after exclusion of g(θ) the second order differ-ential equation for ψ(θ) which is not detailed in /18/. Replacing in it

Φ = dθ/dψ (5.47)

we derive the first order differential equation

(cot 2ψ)dΦ/dψ = 2Φ(µ − 1 + 2µ/Ψ − (1 + 2µ2/Ψ)Φ + µ2Φ2/Ψ) (5.48)

whereΨ = µ + (1 − µ) cos2 2ψ. (5.49)


0

0.2

0.4

0.6

0.78 1.2 1.6 1.96 2.6

3

ψ

0.67 1.33 2.67 θ

Fig. 5.9. Function ψ(θ) at different λ and µ = 0.5

Expression (5.48) should be solved at different Φ(0) = Φo. Then we findfunction θ(ψ) at θ(0) = 0 that corresponds to θ = λ, Φ = 1 at ψ = π/4(Fig. 5.9 for µ = 0.5) and finally we have g(θ) and U(θ).

Here we must notice that equations (5.41), (5.44) give different valuesof g(θ) since we use conditions g(0)/D = 1,U(0)/D = 1. To get the cor-rect answer we recommend the following procedure. We compute g1/D,U1/Daccording to expressions (5.41), (5.44) respectively. Then we find /U2//D bythe first relation (5.41), calculate difference U/D = (U1 − /U2//D) accordingto the second law (5.41). Since g and ψ in (5.45) do not depend on r we canrepresent the normal stresses in form

σrσθ

= ω(t)(A + r−2µ(K(θ) ± gµ cos 2ψ)) (5.50)

where according to the first equilibrium law (2.67)

K(θ) = ((gµ sin 2ψ)′ + 2(1 − µ)gµ cos 2ψ)/2µ.

Using the first integral static equation (3.32) and condition σθ(a, λ) =σr(a, λ) = −q∗ we have

σrσθ

= q∗((gµ(λ) + J(λ))/ sin λ − (a/r)2µ2µ(K(θ) ± gµ(θ) cos 2ψ))/B3,

τ = (q∗/B3)(a/r)2µgµ(θ) sin 2ψ, τe = (q∗/B3)(a/r)2µgµ(θ).(5.51)

HereB3 = (gµ(θ) sin 2ψ)′

∣∣∣θ=λ

− (gµ(λ) + J(λ))/ sin λ

and

J =

λ∫

0

gµ(θ)(sin 2ψ sin θ + 2 cos 2ψ cos θ)dθ.


0

1.5

3

4.5

J1

0.75 1.5 2.25 λ

Fig. 5.10. Dependence of J on λ at µ = 0.5

The maximum τe is at r = a and here we can use the criterion of infiniterate of the biggest elongation ε which gives with consideration of (5.28) andthe second expression (5.51)

ε∗ = 1/α,Ω(t∗) = ((q/B3)max gµ(θ)eα)−m.


At µ = 0 we have from (5.46) expression (4.14) and hence the solution ofSect. 4.2.2. So from (4.15) we find (solid line in Fig. 5.11)

τe = τyi = q∗/n ln(n/(n − 1)). (5.52)

where n is linked with λ by relation in the above mentioned paragraph.At µ = 1 we derive from (5.46), (5.42) and the symmetry condition

g sin 2ψ = 2D sin 2θ (5.53)

and from (5.51) we derive (broken line in Fig. 5.11)

max τe/q∗ = 0.75x1cot λ

λ ≥ π/4λ ≤ π/4 (5.54)

At µ = 0.5 we have from (5.46)

√g sin 2ψ = Hsin θ (5.55)

where H is a constant. Putting (5.55) into (5.42) we receive differentialequation

dθ/dψ = (1 + 2 cot2 2ψ)/(1 + cot θ cot 2ψ). (5.56)

which should be integrated at different Φ(0) = Φo at boundary conditionθ(0) = 0 and it gives values of λ at ψ = π/4. Sokolovski /18/ has represented


0

0.5

1

max τe/q*

30 60 λ°

Fig. 5.11. Dependence max τe(λ) at different µ

the results for λ < π/4. We made the computations for all λ < π (Fig. 5.9).From (5.51), (5.55) and (2.65) we find maximum shearing stress

τe = 2q∗ sin θ√

1 + 1/ tan2 2ψ/B4. (5.57)

HereB4 = (2J1 + λ)/ sin λ − cos λ

and (Fig. 5.10)

J1 =

λ∫

0

(sin 2θ/ tan 2ψ)dθ.

Seeking dτe/dθ = 0 we have with the consideration of (5.56) conditiontan 2ψ = 2 tan θ which gives to (5.57) at θ = λ, a = r

max τe = q∗

√cos2 λ + 4 sin2 λ/B4. (5.58)

Diagram of (5.58) is drawn in Fig. 5.11 by broken-pointed curve.

5.2.4 Penetration of Wedge and Load-Bearing Capacityof Piles Sheet

Putting σr, σθ from (5.50) and τ from (5.45) into (4.20) we receive

P/2l = p∗(1 + a/l) sin λ + ωB5/a2µ (5.59)

where

B5 = ((1 + l/a)1−2µ − 1)(K(λ) sin λ + gµ(λ)(a/l) cos λ − J2)/(1 − 2µ),

J2 =

λ∫

0

((K(θ) + gµ cos 2ψ) cos θ + gµ sin 2ψ sin θ)dθ.


Computing according to (2.65), (5.45) τe we find for its maximum at r = a

max τe = 2(P/2l − p∗(1 + a/l))max gµ(θ)/B5. (5.60)

At a → 0 we have the case of the wedge penetration and at a → ∞, λ → πwe come to the load-bearing capacity of piles sheet. Now we consider theparticular cases.

At µ = 1 we compute from (2.65), (3.31) at C = 0 and (4.14)

max τe = 2(P/2l sin λ − p∗(1 + a/l))xsin2 λcos2 λ/(1 + (4/3) sin2 λ

−(3 − 4 sin2 λ)/(1 + l/a))at π/2≤λ≤3π/4at 3π/4≤λ≤π

which gives at a = 0 and a → ∞, λ → π respectively

max τe = 2(P/2l sin λ − p∗)xsin2 λcos2 λ/(1 + (4/3) sin2 λ,

at π/2≤λ≤3π/4at 3π/4≤λ≤π

Pyi = 2(p∗b + τyil) (5.61)

where the last member in the second expression is added from mechanicalconsiderations.

At µ = 0.5 we receive with (4.20), (5.51) and (5.56).

max τe = (P/2l − p∗(1 + a/l) sin λ)√

cos2 λ + 4 sin2 λ/2((a/l) ln(l/a + 1) sin 2λ− (λ + J1)). (5.62)

For the cases a = 0 and a → ∞, λ → π we have respectively

max τe = (P/2l − p∗ sin λ)√

cos2 λ + 4 sin2 λ/2(λ + 2J1),max τe = (P/2 − p∗b)/2(π + 2J(π))

(5.63)

and the expression like the second one (5.61) and similar to (5.29) as well asaccording to the criterion dγ/dt → ∞

Pu = 2(τtl + p∗b); ε∗ = 1/α,Ω(t∗) = (αe2max τe)−2 (5.64)

For the last relations the following constitutive equation is used (seerheological law (5.28) at µ = 1/2.)

Ω(t)(2max τe)2 = εe−αε

5.2.5 Wedge Under Bending Moment in its Apex

We will seek a rigorous solution of this task (Fig. 5.12) at σθ = 0. Then wehave from expressions (2.67) at τrθ ≡ τ equations /18/

∂(rσr)/∂r + ∂τ/∂θ = 0, ∂(r2τ)/∂r = 0 (5.65)


M

x

y

r

Fig. 5.12. Wedge under moment in its apex

that give solutions in form

τ = f(θ)/r2, σr = f ′(θ)/r2 (5.66)

where f is a function of θ only. Boundary conditions for τ are

τ(±λ) = 0. (5.67)

Above that on the neutral axis (θ = 0) as in the beam at bending similar tothe problem of the slope under one-sided load we have σr = 0.

Components of main vector in any cross-section r = constant

∑X = r

λ∫

−λ

(σr sin θ + τ cos θ)dθ,∑

Y = r

λ∫

−λ

(σr cos θ + τ sin θ)dθ

with a help of (5.65), (5.66) can be rewritten as

λ∫

λ

(f sin θ)′dθ,

∫ λ

λ(f cos θ)′dθ

and according to (5.66), (5.67) they are equal to zero.Now we use representations similar to (5.18)

εr = 0.5g(θ)r−2m cos 2ψ, γ = g(θ)r−2m sin 2ψ (5.68)

where g(θ) = γmr2m and according to (5.17)

σr = 2ω(t)(gµ/r2) cos 2ψ, τ = ω(t)(gµ/r2) sin 2ψ. (5.69)


Putting (5.68) into the first static equation (5.65) we find

(gµ sin 2ψ)′ − 2gµ cos 2ψ = 0. (5.70)

In the same manner we derive from (5.68), (2.71)

(g cos 2ψ)′′ + 4m(1 − m)g cos 2ψ + 2(2m − 1)(g sin 2ψ)′ = 0. (5.71)

At µ = m = 1 we have from (5.70), (5.71) equation

(g cos 2ψ)′′ + 4g cos 2ψ = 0

with obvious solution

g cos 2ψ = Ccos 2θ + D sin 2θ.

Condition above σr(r, 0) = 0 gives C = 0 and from (5.65), (5.66) as well asobvious integral static law

M = −2

λ∫

0

τ(θ)r2dθ (5.72)

we compute

σr = 2(M/r2B6) sin 2θ, τ = (M/r2B6)(cos 2λ − cos 2θ) (5.73)

whereB6 = sin 2λ − 2λ cos 2λ. (5.74)

Now (2.65) gives

τe = M(1 + cos2 2λ − 2 cos 2θ cos 2λ)0.5/r2B6. (5.75)

Condition dτe/dθ = 0 leads to equality sin 2θ = 0 with solutions θ = 0 andθ = π/2. Investigations show that max τe is at θ = 0 in form

maxτe = 2M(sin2 λ)/B6. (5.76)

Diagram max τe(λ) is drawn by solid line in Fig. 5.13. From relations (5.60),(5.73) we derive expression (5.22) with sign minus and hence, solid lines inFig. 5.8 reflected relatively to axis θ.

In the general case we derive from (5.70), (5.71) after an exclusion of g thesecond order differential equation

tan 2ψψ′′ − 2(1/µ − 1 + 2/Ψ)ψ′2 + 2(1 + 2/µΨ)ψ′ − 2/µΨ = 0 (5.77)

where Ψ is given by relation (5.49). Now we suppose

dθ/dψ = Θ (5.78)


0

1

2

3

max τer2/M

45 60 90 λ°


and (5.77) becomes

(tan 2ψ)dΘ/Θdψ + 2(1/µ− 1 + 2/Ψ)− 2(1 + 2/µΨ)Θ + 2Θ2/µΨ = 0. (5.79)

According to (5.70) equation (5.79) must be solved at different Θ(-π/4) = Θo.Then we integrate (5.78) with border demand θ(-π/4) = 0 (see Appendix G).

To find max τe we write from (5.70) with consideration of (5.78) as

g(θ)/g(0) = exp((2/µ)

θ∫

0

(1 − 1/Θ) cot 2ψdθ). (5.80)

Now from (5.69), (5.72) we find at θ = ß

max τe = M/2Jr2 (5.81)

where

J =

λ∫

0

(g(θ)/g(0))µ sin 2ψdθ. (5.82)

A very simple solution takes place at µ = 0 when we have from (5.70) withconsideration of boundary condition above relation

ψ = θ − π/4

and from (5.69) we find expression

τ = −(ω(t)/r2) cos 2θ.

Putting it into (5.72) we receive at θ = 0 (broken line in Fig. 5.13)

max τe = M/r2 sin 2λ. (5.83)

We can see that this value coincides with (5.78) only at λ = π/4 and at biggervalues of λ it is above the curve for m = 1.


5.2.6 Load-Bearing Capacity of Sliding Supports

In order to improve conditions of sluice and lock exploitation the inconvenientwheels and rollers are often replaced by sliding supports. The latter usuallyconsists /27/ of embraced in metal holder 1 (Fig. 5.14) polymer skid 2 alongwhich steel rail 3 moves. The structure hinders from longitudinal displacementof the skid and because of that the greatest interest has an appreciation of itsresistance to compression.

A rigorous solution of this task with consideration of complex boundaryconditions and peculiarities of the polymer’s mechanical behaviour is hardlypossible. At the same time constantly widening use of such structures in dif-ferent branches of industry compels to seek simple and convenient for prac-tice approaches. Because of that we give here the approximate solution basedon results of study of this problem for simple media and some kinematichypothesis.

Here we use the first expression (5.17) as

σ1 − σ3 = 4ω(t)εµ. (5.84)

On the base of applied plasticity theory /10/ (see also Sect. 4.2.2) and withconsideration of Coulomb’s friction law the following expressions for parts Iand II of the skid (Fig. 5.14) are formulated (see Appedix H)

4ω(t)(εxI)µ = σo + /N/f/h(e2fa/h − 1),

4ω(t)(εyII)µ = −fhσo/(efh/(b−a) − 1)(b − a).(5.85)

Here f is friction coefficient, dimensions a, b, h are shown in Fig. 5.14 andvalue of σo in the second expression (5.85) is determined from integral staticequation. Excluding from (5.85) σo and using the constant volume demand inform

εxI = (β − 1)εyII (5.86)

where β = b/a we derive

1

II

a abb

y

2h

x

N dh3

III

Fig. 5.14. Sliding support


4ω(t)(εxI)µ = /N/f/h(e2fh/a−1)(1+(b−a)(efh/(b−a)−1)/fh(β−1)µ). (5.87)

When εxI is known the most interesting value of track’s depth dh can be foundfrom (5.86) as

dh = hεxIβ/(β − 1). (5.88)

The approximate solution (5.87), (5.88) must be compared with similarresults for ideal bodies according to the Hooke’s law when µ = 1, ω = G, andperfect plasticity with µ = 0, σ1 − σ3 = σyi The comparison will be made forsmooth surfaces contact of the skid with the holder and the rail. So, at f = 0we have from (5.87), (5.88)

Gdh/N = hβ/8b

(solid lines 1, 2 in Fig. 5.15 for b/h = 1, 2 respectively) and

aσyi/N = 1/4

(solid straight line in Fig. 5.16).The solution of the problem for linear material is given in the form

dh = C − /N/F(ϕ,dn(K/β))/4GK1 (5.89)

whereϕ = sin−1((sn(Kx/b)/sn(K/β)),

sn, dn – sine and delta of amplitude (the Jakobi’s elliptic functions), K, K1 –full elliptic integrals of the first kind with moduli k and ksn(K/β) respectively.The dimensions are linked by relation

K(k)/K(k′) = b/h. (5.90)

Here k′ =√

1 − k2 – modulus additional to k.

0

0.08

0.16

0.24

Gdh/N

1

1

2

2

3

Fig. 5.15. Dependence of dh on β for elastic material


0

0.08

0.16

0.24

1 2 3

Fig. 5.16. Compression of ideal plastic layer

On the base of theoretical-experimental investigation R. Hill /28/ gavethe dependence of punch pressure on the thickness of a layer of ideal plasticmaterial that in terms of our work is drawn by broken line in Fig. 5.16. Atβ > 2 it is near to that of expressions (4.39).

From Figs. 5.15, 5.16 we can see that at b/h near 2 and β > 2 the proposedhere simple solution agrees well enough with similar results for ideal bodies.As in practice dimensions of the skid are taken as b near to 2h and β = 1.5. . . 3the represented relations can be used for applications. To receive calculationexpressions we exclude from (5.87). (5.88) εxI that gives

dh = Ω(t)(N/4)mR (5.91)

where

R = βh1−m/((e2fa/h − 1)(1 + (b − a)(efh/(b−a) − 1)/fh(β − 1)µ)m(β − 1).

Relation (5.91) has the same properties as similar creep laws for tensionand in complex stress state and it can serve for computation of dh accordingto the test data in compression.

5.2.7 Propagation of Cracks and Plastic Zones near Punch Edges

General Relations

Similar to Sect. 5.1.1 we take the potential function as

Φ = Krsf(θ)

We find the stresses according to (2.75) and strains by (2.61). Since sum ofproducts σijεij is proportional to r−1 /21/ we have s = (2m + 1)/(m + 1),


σr = Kr−1/(m+1)(sf + f ′′), σθ = Kr−1/(m+1)s(s− 1)f, τrθ = Kr−1/(m+1)(1− s)f ′

(5.92)and

εθ = −0.5Ω1(t)Kmr−m/(m+1)Fm−1(f ′′ + s(2 − s)f),γrθ = 2Ω1(t)(1 − s)Kmr−m/(m+1)Fm−1f ′.

(5.93)

Here Ω1 is proportional to Ω, x, y in (2.61) must be replaced by r, θ respec-tively and

F = ((f ′′ + f(1 + 2m)/(m + 1)2)2 + (2mf ′/(m + 1))2)0.5. (5.94)

Putting εθ, γrθ into compatibility law (2.71) we get a very complex non-lineardifferential equation of the fourth order which must be solved together withthe consequent boundary conditions.

Crack in Tension and Pressure of Punch

The border demands for these problems are respectively: f(0) = 1, f ′(0) =f ′(π) = f(π) = 0 and f(π) = −1, f ′(0) = f ′(π) = f(0) = 0.

Since for the tasks f ′(0) = f ′(π) = 0, in perfect plastic solution f ′′′(0),f ′′′(π) and this value at π/4 < θ < 3π/4 is also equal to zero as well as in con-sequent elastic relations (Sects. 3.2.7 and 3.2.9 where f(θ) is proportional tocos3 θ and sin3 θ respectively) f ′′′(0) = 0 we suppose f ′ = f ′′′ = 0 everywherein the zones mentioned above and derive equation (see more precise option inAppendix I)

fIv + (3m + 7)f ′′/(1 + m)2 + (2m + 1)(2 + m)f/(m + 1)4 = 0 (5.95)

with obvious solution

f = A1 cos β1θ + A2 sin β1θ + A3 cos β2θ + A4 sin β2θ. (5.96)

Hereβ1

β2= ((−3m − 7 ±

√m2 + 22m + 41)/2(1 + m)2)0.5,

and Aj (j = 1. . . 4) can be found from the boundary conditions (it is inter-esting to notice that at m = 1 this approximate solution coincides with therigorous one – see consequent relations in Sect. 3.2.8). Then we compute f ′,f ′′ and the stresses according to (5.92). Calculations show that their diagramsfor m = 3, m = 15 are near to ones in Fig. 4.19 for elastic and perfect plasticbodies respectively.

In the general case we can find K according to the condition /21/ of theindependence of integral J = dW/dl on the properties of the material. Here

dW =

dl∫

0

σθ(x, 0)uθ(dl − x, π)dx. (5.97)


Putting σθ, uθ from (3.90), (3.91) into (5.97) we find for m = 1

J1 = (1 + κ)πσ2l/8G. (5.98)

To get J in the general case we use relations (5.93), (5.92) for εr, εθ, γrθ, σθand compute uθ from (2.69). The latter gives

uθ = Ω1(t)Kmr1/(m+1)Fm−1((6m+1)f ′+(m+1)2((m−1)F′(f ′′+fs(2−s))/F+f ′′′))/4m

where

F′/F = F−2((f ′′ + fs(2 − s))(f ′′′ + f‘s(2 − s)) + 4(1 − s)f ′f ′′)).

Now from (5.97), (5.92) and expression for uθ above we derive J for any m

J = 0.5Km+1m(2m + 1)/f ′′(π)/m−1f ′′′(π)

1∫

0

((1 − ξ)/ξ)1/(1+m)dξ.

Using equation J = J1 we have

K = ((κ + 1)πσ2l/4GΩ1(t)I1(m))1/(1+m).

Here

I1(m) = (2m+1)(m+1)/f ′′(π)/m−1f ′′′(π)Γ((2m+1)/(m+1))Γ((2+m)/(1+m))

and Γ() is a Gamma-function. Lastly, according to condition τe = constantwe compute equation

2r(2τe)m+1GΩ1I1(m)/σ2(κ + 1)πl = Fm+1 (5.99)

which determines yielding zone near the crack edges.

Transversal Shear

It can be considered in a similar manner. As in this problem f(0) = f(π)= 0, we neglect here f and f′′ versus f′ and find from (5.94)

F = 2mf ′/(m + 1).

Compatibility law (2.71) gives equation

fIv + (5m2 + 4m + 1)f ′′/(m + 1)2 + m(2m + 1)(m + 2)f/(m + 1)4 = 0

with solution (5.96) in which

β1

β2= ((−(5m2 + 4m + 1) ±

√25m4 + 32m3 + 6m2 + 1)/2)0.5/(m + 1).


For the crack and punch we have two other border demands as f ′(π) = 0,f ′(0) = 1 and f ′(π) = 1, f ′(0) = 0 respectively. It is interesting to notice thatonce again the approximate solution gives at m = 1 the rigorous results.

Now we find f′, f′′(see calculations in Appendix J) and according to (5.92) –stresses σθ, σr, τrθ, τe., strain εθ, displacement ur, integral J and factor K as

Km+1 = (κ + 1)πτ2l/2I(m)/f ′′(π)/m

where I(m) is the same as in Sect. 5.2.1. Condition τe= constant gives equationfor r in form

2r(2τe)m+1GΩ(t)/τ2(κ + 1)πl = Fm+1/I(m)/f ′′(π)/m. (5.100)

Diagrams σθ/σyi, σr/σyi, τrθ/σyi are given in Fig. 3.11 by the same lines asfor m = 1 but with index 0. From the figure we can see that with the growthof m the distribution of stresses changes very strongly. In the same mannerthe problem of the punch horizontal movement can be considered. The curvesfor the stresses can be received by reflection of the previous ones relatively toaxis θ = π/2.

5.3 Axisymmetric Problem

5.3.1 Generalization of Boussinesq’s Solution

As in Sect. 3.3.2 we suppose for incompressible material (ν = 0.5) σχ = σθ =τρχ = 0, and from the first static equation (2.77) as well as from rheologicallaw (1.29) at α = 0 we have for stress and strain following relations

σρ = f(χ)/ρ2, ερ = g(χ)Ω(t)ρ−2m (5.101)

where g = f m. Since for this case in (2.79) εχ = εθ we find easily

uχ = U(ρ) sin χ,uρ = ϕ(χ) + g(χ)Ω(t)ρ1−2m/(1 − 2m). (5.102)

Using condition ερ = −2εχ we derive

0.5Ω(t)(3 − 2m)g(χ)ρ1−2m/(1 − 2m) + ϕ(χ) + U(ρ) cos χ = 0. (5.103)

Putting uρ, uχ from (5.102) into condition γρχ = 0 we determine

ϕ′(χ) + g′(χ)Ω(t)ρ1−2m/(1 − 2m) + ρ2(sin χ)∂(U(ρ)/ρ)/∂ρ. (5.104)

Excluding ϕ′(χ) from (5.103), (5.104) we obtain the expression in whichboth parts must be equal to the same constant, say n, since each of themdepends only on one variable (neglecting t as a parameter) in form

Ω(t)g′(χ)/2 sin χ = ρ2mdU(ρ)/dρ = −n


P

Z

Fig. 5.17. Computation of constant n

with obvious solutions

f(χ) = Ω−µ(C + 2n cos χ)µ,U(ρ) = D − nρ1−2m/(1 − 2m). (5.105)

Since at χ = π/2 we have σρ = 0 we must put in the first (5.105) C = 0 andconstant n should be found from condition (Fig. 5.17)

P = −2

λ∫

0

σρρ2 sin χ cos χdχ. (5.106)

Putting here σρ from (5.101) we find after calculations

σρ = −P(µ + 2) cosµ χ/2π(1 − cosµ+2 λ)ρ2. (5.107)

Taking in the second relation (5.105) D = 0 we get the displacement as

uχ = −Ω(t)(P(µ + 2)/2π(1 − cos2+µ λ))mρ1−2m(sin χ)/(1 − 2m). (5.108)

The most interesting case takes place at λ = π/2 when we receive fromexpressions (5.107), (5.108)

σρ = −P(2 + µ)(cosµ χ)/2πρ2,

uχ = −Ω(t)(P(2 + µ)/2π)mρ1−2m(sin χ)/(1 − 2m).(5.109)

It is easy to notice that the highest value of σρ at ρ = constant is on the lineχ = 0. It is not difficult to find that there stress σρ at m = 1 is 1.5 times morethan at µ = 0. The biggest value of uχ is at χ = π/2 but its dependence onm is more complex. However the second relation (5.109) allows to calculatethe displacements in some distance from the structure loaded by forces witha resultant P.


0 0.2 0.4

2

4

z/a

Fig. 5.18. Comparison of stress distribution

To appreciate a practical meaning of the results we compare for m = 1the distribution of stress σz on axis z for the concentrated force P = qa2

and for the circular punch of radius a when we have from (5.109) and (3.122)respectively

/σz//q = 3(a/z)2/2π, σz/q = 1 − (1 + (a/z)2)−3/2. (5.110)

From Fig. 5.18 where by solid and broken lines diagrams /σz/(z) are shownwe can see that at z/a > 3 the simplest solution for concentrated force can beused. Since at µ < 1 a distribution of stresses becomes more even we canexpect better coincidence of similar curves with the growth of a non-linearity.

It is interesting to notice that according to Figs. 5.18, 5.4 vertical stress inaxisymmetric problem is approximately twice less than in the plane one. Thisexplains higher load-bearing capacity of compact foundations.

5.3.2 Flow of Material within Cone

Common Equations

We solve this problem at the same suppositions as that in Sect. 4.3.3 From(2.79) at uχ = 0 we compute

ερ = −2U/ρ3, εθ = εχ = U/ρ3,

γρχ ≡ γ = dU/ρ3dχ, γm = g/ρ3(5.111)

where U = U(χ) andg(χ) =

√9U2 + U′2.

Similar to (5.18) and (5.68) we use representations

εχ = −g(cos 2ψ)/3ρ3, γρχ = g(sin 2ψ)/ρ3 (5.112)


putting which into the first law (2.82) we have equations

(g cos 2ψ)′ + 3g sin 2ψ = 0,

dg/gdχ = 2(dψ/dχ − 3/2) tan 2ψ.(5.113)

The latter gives boundary condition dψ/dχ = 3/2 at ψ = π/4. From expres-sions for strains above we can also find

dln/U//dχ = −3 tan 2ψ, g(χ) = −3U/ cos 2ψ, (5.114)

From (5.17) and (5.112) we derive representations

τρχ = τ = ω(t)ρ−3µgµ sin 2ψ,σρσχ = ω(t)(C + ρ−3µ(K+2

−1x2gµ(cos 2ψ)/3))(5.115)

where C is a constant and function K(θ) can be found from the first staticequation (2.77) as follows

3µK = (gµ sin 2ψ)′ + cot χ(gµ sin 2ψ) + 4(1 − µ)gµ cos 2ψ. (5.116)

Putting (3.115) into (2.78) we derive

(gµ sin 2ψ)′′ + (gµ sin 2ψ)′ cot χ + (9µ(1 − µ) − 1/ sin2 χ)gµ sin 2ψ

+ 2(2 − 3µ)(gµ cos 2ψ)′ = 0.(5.117)

Combining (5.113), (5.117) we have at Ψ according to (5.49) two differentialequations

Θ = dχ/dψ, (5.118)

(cot 2ψ)dΘ/dχ − 2(µ − 1 + 2µ/Ψ) + Θ((6µ2 + 3µ + 4(1 − µ) cos2 2ψ)/Ψ

− cot χ cot 2ψ) − 3Θ2(3µ2 − (µ + µ tan 2ψ cot χ + 1/3 sin2 χ) cos2 2ψ)/2Ψ = 0(5.119)

the second of which should be solved at different Θo = Θ. Then we integrate(3.118) at border demand χ(0) = 0. The searched function must also satisfycondition χ = λ at ψ = π/4. Now we receive from (5.113), (5.114), (2.65)U(χ), g(χ) and τe.

Putting stress σρ from (5.115) into integral static equation (4.105) we findat −q∗ = σρ(a, λ) = σχ(a, λ) expression for max τe as

max τe = 3µq∗maxgµ(χ)/((gµ(θ) sin 2ψ)′∣∣∣χ=λ

− gµ(λ) cot λ − 2J3/ sin2 λ)

where

J3 =

λ∫

0

gµ(θ)(sin 2ψ sin2 χ + 2 cos 2ψ sin 2χ)dχ.

Then the criteria max τe = τu and dγm/dt → ∞ must be used as before. Forthe latter we have

ε∗ = 1/α,Ω(t∗) = (αe2max τe)−m.



At µ = 0 we have the solution of Sect. 4.3.3.If µ = 1 we compute from (5.113), (5.117) equation

(g sin 2ψ)′′ + (g sin 2ψ)′ cot χ + (6 − 1/ sin2 χ)g sin 2ψ = 0

with obvious solutiong sin 2ψ = 2D sin 2χ (5.120)

where D is a constant. Then from (5.112)

g cos 2ψ = 3D(cos 2χ − cos 2λ). (5.121)

From (5.120), (5.121) we receive

tan 2ψ = 2(sin 2χ)/3(cos 2χ − cos 2λ)

Diagrams ψ(χ) at different λ according to this relation are drawn in Fig. 5.19.Similar to the general case we have ultimate condition as

max τe = q∗x12/3 tan λ (λ>

<33.7) (5.122)

At µ = 2/3 we calculate from (5.117) equation

(g2/3 sin 2ψ)′′ + (g2/3 sin 2ψ)′ cot χ + (2 − 1/ sin2 χ)g2/3 sin 2ψ = 0

with obvious solution solution

g2/3 sin 2ψ = Hsin χ (5.123)

where H is a constant. Putting (5.123) into (5.113) we derive differentialequation of the first order

0

15

30

30 60

Fig. 5.19. Dependence ψ(χ) at µ = 1


0 60 120

2.3 3.051.280.58

1

15

30

4 8 10

χo

ψo

Fig. 5.20. Diagrams ψ(χ) at µ = 2/3 and different Θo

dχ/dψ = 2(2 + 3 cot2 2ψ)/3(2 + cot χ cot 2ψ) (5.124)

that should be integrated at different Θ(0) = Θo. Diagrams χ(ψ) at Θo-valuesin the curve’s middles and λ at their tops are given in Fig. 5.20.

Putting σρ from (5.113) into (4.105)we find C and from conditiondτe/dχ = 0 with consideration of (5.124) – equality tan 2ψ = 3 tan χ whichgives to max τe (it increases with a growth of χ) value

max τe = q∗ sin2 λ√

cos2 λ + 9 sin2 λ/(3 cos λ − cos3 λ − 2 − 12J4).

Here as before −q∗ = σρ(a, λ) = σχ(a, λ) and

J4 =

λ∫

0

(sin2 χ cos χ)(tan 2ψ)−1dχ.

Diagrams J4(λ) and max τe(λ) are shown in Figs. 5.21, 5.22 respectively. Thebroken line in the latter picture refers to the case µ = 1 (computations forµ = 1/3 see in Appendix K-interrupted by points curve in the figure) andpointed line refers to solution (4.106).

5.3.3 Cone Penetration and Load-Bearing Capacityof Circular Pile

Here common relations (5.111). . . (5.119) are valid. We put stresses accordingto (5.115) into integral static equations (4.107). to detail the constants andaccording to (2.65) we compute max τe at a = ρ as


0

0.4

0.8

1.2

J4

45 90 135

Fig. 5.21. Diagram J4(λ) for µ = 2/3

0

1

2

30 60


max τe = 3µ(P/π − p∗(a + 1)2 sin2 λ)(maxgµ(χ))/a(a(2(gµ(χ) sin 2ψ)′∣∣∣

χ=λsin2 λ

+ gµ(λ)(1 + 3µ) sin 2λ((1 + l/a)2−3µ − 1)/(2 − 3µ)

− l(gµ(λ) sin 2λ + 2J5)(2 + l/a)) (5.125)

where p∗ is the strength of soil in a massif at compression and

J5 =

λ∫

0

gµ(χ)((1 + 3µ) sin 2ψ sin2 χ + 2 cos 2ψ sin 2χ)dχ.

At λ → π, a → ∞ we find for a circular pile

max τe = (P/π − p∗b2)maxgµ(χ)/l(bgµ(λ) + J5(λ))(2 + l/a). (5.126)

In the same manner we consider the particular cases and consequently forµ = 0 we receive from (5.117), equation (4.103) and hence the solution ofSect. 4.3.4.


At µ = 1 we have

max τe = 4.5(P/π − p∗(a + 1)2 sin2 λ)x12/3 tan λ/la(2(5 − 6 sin2 λ)/(1 + l/a)

− (2 + 3 sin2 λ)(2 + l/a))λ<146

λ<146

and for the pile the yielding (the first ultimate) load is

Pyi = πb(2τyil + p∗b). (5.127)

We can see that this result has obvious structure and coincides with approxi-mate relation (4.110) for ideal plasticity.

Similarly we compute for µ = 2/3

max τe = (P/π − p∗(a + 1)2 sin2 λ)√

cos2 λ + 9 sin2 λ/6a(2a cos λ sin2 λ ln(1 + l/a)

− l(2 + l/a)(1 − cos λ + 2J4)).

Here J4 is given in Sect. 5.3.2. For the circular pile this relation predicts bigvalues of ultimate load and so we can take in the safety side

Pu = πb(p ∗ b + 2τul). (5.128)

5.3.4 Fracture of Thick-Walled Elements Due to Damage

Stretched Plate with Hole

We consider plate of thickness h with axes r, θ, z (Fig. 5.23) and use theTresca-Saint-Venant hypotheses. Since here σθ > σr > σz = 0 we have εr = 0and from (2.32) at α = 0, σeq = 2τe = σθ

ε ≡ εθ = 3Ω(t)σθm/4

σθ = (4/3Ω)µεµ (5.129)

where ε = u/r and radial displacement u depends only on t.Putting (5.129) into the static equation of this task

hσθ = d(hrσr)/dr, (5.130)

integrating it at h = constant as well as at boundary conditions σr(a) = 0,σr(b) = p and excluding factor (4u/3Ω)µ we receive with the help of (5.129)

σθ = p(1 − µ)(b/r)µ/(1 − βµ−1) (5.131)

where β = b/a. Putting σθ into (2.66) we find for the dangerous (internal)surface

e−αεε = 3β(1 − µ)mΩ(t)pm(1 − βµ−1)m/4. (5.132)

Applying to (5.132) criterion dε/dt→ ∞ we have


a

b

r up

Fig. 5.23. Stretched plate

ε∗ = 1/α,pmΩ(t∗) = 4(1 − βµ−1)m/3β(1 − µ)mαe. (5.133)

When the influence of time is negligible we compute from (5.133) atΩ = constant critical load

p∗ = (4/3)µ(1 − βµ−1)/(1 − µ)(αβΩe)µ. (5.134)

At small µ that value must be compared to ultimate load pu which followsfrom (5.131) at µ → 0 as

pu = σyi(1 − 1/β)

where σyi is a yielding point at an axial tension or compression and the small-est value should be taken. At m near to unity we must compare p∗ withyielding load which follows from (5.131) at m = 1 in form:

pyi = (σyi/β) ln β

and the consequent choice should be made.

Sphere

For a sphere under internal q and external p pressures (Fig. 3.23) we denotethe radial displacement also as u and according to relations (2.80) we computeεθ = u/ρ, ερ= du/dρ and from the constant volume demand (2.81) we find

u = C/ρ2, ερ = −2C/ρ3, εθ = C/ρ3 = u/ρ (5.135)

where constant C is to be established from boundary conditions. Now from(2.32) at α = 0 and σeq = σθ − σρ we deduce


εθ = Ω(t)(σθ − σρ)m/2

or with consideration of (5.135)

σθ − σρ = (2C/Ωρ3)µ. (5.136)

Putting (5.136) into static equation (2.80) we get after integration at borderdemands σρ(b) = −p, σρ(a) = −q and exclusion of constants

σθ − σρ = 3(q − p)µ(b/ρ)3µ/2(β3µ − 1). (5.137)

Now we use constitutive law (2.32) which for our structure is

e−αεε = 0.5Ω(t)(σθ − σρ)m (5.138)

where ε = εθ. Using here σθ − σρ from (5.136) and criterion dε/dt→ ∞ wededuce

ε∗ = 1/α, (q − p)mΩ(t) = 2(2m/3)m(1 − β−3µ)m/αe. (5.139)

When the influence of time is not high critical difference of the pressures atΩ = constant can be got

(q − p)∗ = 21+µm(1 − β−3µ)/3(αΩe)µ. (5.140)

At small µ this value should be compared with (q − p)u according to (4.97)and the smaller one must be taken. Similar choice have to be fulfilled between(q − p)∗ and (q − p)yi given by (4.94) at µ near unity.

Cylinder

In an analogous way the fracture of a thick-walled tube can be studied. From(2.32) at α = 0, εx = 0, εθ ≡ ε and σeq = σθ − σr we have

ε = (3/4)Ω(t)(σθ − σr)m (5.141)

and providing the procedure above for the disk and the sphere we find/17, 27/

σθ − σr = 2µ(q − p)(b/r)2µ/(β2µ − 1). (5.142)

Equation (2.32) for this structure is

e−αεε = 3Ω(t)(σθ − σr)m/4. (5.143)


Using here expression (5.142) at r = a and the criterion dε/dt → ∞ we derive

ε∗ = 1/α, (q − p)mΩ(t∗) = 4(m/2)m(1 − β−2µ)m/3αe. (5.144)

When influence of time is negligible we can find as before critical differenceof pressures as

(q − p)∗ = (4/3)µm(1 − β−2µ)/2(αΩe)µ

and again for µ near to zero this value must be compared with (q − p)uaccording to (4.99) and smaller one have to be taken. The similar choiceshould be made between (q − p)∗ and (q − p)yi from relation (4.98) at mnear unity.

From Fig. 5.24 where at α = 1, m = 1 and m = 2 by solid and broken lines1, 2, 3 for plate, sphere and cylinder curves t∗(β) are represented respectivelywe can see that the critical time for a tube is less than the consequent one forthe sphere and higher that of the plate.

Cone

We consider this task at the same suppositions as in Sect. 3.3.1 and Sect. 4.3.1.Using the scheme above, relations for strains (3.117) and stresses (3.116) aswell as law (5.141) at σr → σχ we find

σθ − σχ = (q − p) sinµ χ/J6 sin2µ χ (5.145)

where

J6 =

λ∫

ψ

(cos1+µ χ/ sin2µ+1 χ)dχ. (5.146)

10

0.5

Ω * (q−p)m

4

3

1

2

2

β

Fig. 5.24. Dependence of t∗ on β and m


The computations for m = 1 when J6 = A/2 from Sect. 3.3.1 at λ = π/3,ψ = π/6 show that integral J6 can be easily calculated. For example at m = 2and the shown meanings of λ, ψ its value is 0.9.

In order to appreciate the moment of fracture we put (5.145) into (5.143)and use criterion dε/dt → ∞ when we have for a dangerous (internal) surface

ε∗ = 1/α,Ω(t∗)(q − p)m = 4(J6)m(sin2 ψ)/3eα cos ψ. (5.147)

If the influence of time is negligible we derive from (5.147) at Ω = constant

(q − p)∗ = (4/3)µJ6(sin2 ψ/αeΩ cos ψ)µ. (5.148)

Once more for small µ this value must be compared with (q − p)u accord-ing to (4.100) and smaller one should be taken. At µ near to unity (q − p)∗must be compared to (q − p)yi from Sect. 4.3.1 and similar choice shouldbe made.

Conclusion

The results of the solutions of this paragraph can be used for a prediction ofa failure not only of similar structures but also of the voids of different formand dimension in soil massifs.

6

Ultimate State of Structures at Finite Strains

6.1 Use of Hoff’s Method

6.1.1 Tension of Elements Under Hydrostatic Pressure

This approach takes as the moment of a fracture time t∞ when the structures’dimensions become infinite. We consider as the first example a plate in tensionby stresses p under hydrostatic pressure q (Fig. 6.1). Since here σ1 = σ2 = p,σ3 = −q we have from (2.31)

dε/dt = 0.5B(po)m(eε + κo)m (6.1)

where ε = ε1, κo = q/po. and according to (1.42) p = poeε The integration of(6.1) in limits 0 ≤ ε ≤ ∞, 0 ≤ t ≤ t∞ gives

B(po)mt∞ = 2(ln(1+κo)+(m−1)!

m−1∑i=1

(−1)i(1− i/(1+κo)i)/i!i(m−1− i)!)/(κo)

m.

From Fig. 6.2 where for some κo curves tu(µ) according to the latter expressionare given by broken lines we can see that tu ≡ t∞ diminishes with an increaseof hydrostatic component.

In a similar way the fracture time can be found for a bar in ten-sion by stresses q under hydrostatic pressure p (Fig. 6.3). In this case /17/σ1 = q, σ2 = σ3 = −p and hence in (2.31)

S1 = 2(p + q)/3, σeq = p + q.

Comparing this data to the previous ones we can see that rate dε/dt in thelatter problem is twice of that for the plate. Hence t∞ for the bar is one halfof that in the plate case.

162 6 Ultimate State of Structures at Finite Strains

q

qp

Fig. 6.1. Stretched plate under hydrostatic pressure

1.5

Bpom tcr

1

0.5

0 0.25 0.5 0.75

21

02

1

0

µ

Fig. 6.2. Dependence of ultimate time tcr on κo and µ

q

q

p

Fig. 6.3. Bar in tension under hydrostatic pressure

6.1.2 Fracture Time of Axisymmetrically Stretched Plate

In order to integrate differential equation (5.130) in the range of finite strainswe take according to the condition of Tresca-Saint-Venant in Sect. 5.3.4r = ro + u(t). We replace strains and displacements by their rates and rewrite

6.1 Use of Hoff’s Method 163

(5.130) with consideration of (5.129) at B instead of Ω (see (2.31)) anddε/dt = dr/rdt as

d(rohoσr)/dro = (4/3B)µroho(ro + u)−1−µ(du/dt)µ. (6.2)

Integration of (6.2) at boundary conditions σr(ao) = 0, σr(bo) = p gives

hbbop = (4/3B)µ(du/dt)µ

bo∫

ao

roho(ro)(ro + u)−1−µdro. (6.3)

Here hb = ho(bo). From (6.3) the fracture time can be found. Particularly atho = constant and ho = hbbo/ro we derive after transformations

B(bop)mt∞ = (4/3)

∞∫

0

(((bo + u)1−µ − (ao + u)1−µ)/(1 − µ)

+ u((bo + u)−µ − (ao + u)−µ)/µ)mdu,

Bpmt∞ = (4/3)mm

∞∫

0

((ao + u)−µ − (bo + u)−µ)mdu.

(6.4)

For any m integrals in (6.4) can be computed If e.g. m = 1 we haverespectively at β = bo/ao

Bpt∞ = 4(1 − 1/β)/3,Bpt∞ = (4/3) ln β (6.5)

and from Fig. 6.4 where broken lines 0, 1 are drawn according to (6.5) we cansee that the curved profile has higher critical time. Broken 2 and interruptedby points 1 lines refer to the cases m = 2,ho = hbbo/ro and ho = constantwhen we derive from (6.4)

1

1

2

4

Bpm tcr

2

2

21

0

1

1

β

Fig. 6.4. Dependence of ultimate time tcr on β and m for stretched plate


Bp2t∞ = (16/3) ln((1 +√

β)4/16β),

Bp2t∞ = 16((3(1 + β2)/2 + 2β) − 2(1 + β)√

β + 2(ln((1 +√

β)/2)

+ β2 ln((1 + 1/√

β)/2))/3β2.

6.1.3 Thick-Walled Elements Under Internaland External Pressures

We begin with a sphere and replace εθ, u in (5.135) by their rates dεθ/dt,V. Then we suppose in (5.136) Ω(t) = Bt. According to definition β = b/awe have

dβ/dt = (db/dt − βda/dt)/a (6.6)

where

db/dt = V(b) = bdεθ(b)/dt,da/dt = V(a) = adεθ(a)/dt.

Using (5.136), (5.137) and (6.6) we derive after integration

B(q − p)mt∞ = 2(2m/3)mβ0∫

1

((β3µ − 1)m/β(β3 − 1))dβ. (6.7)

Here βo = bo/ao and critical time is equal to t∞. Diagram tcr(βo) according to(6.7) at m = 1 is given in Fig. 6.5 by broken line 0. The curves for m = 2,m = 3go higher. For them

B(q − p)2t∞ = 128(ln((βo3/2 + 2)/2βo

3/4))/27,

B(q − p)3t∞ = 16(ln βo + 2√

3 tan−1((1 − 1/βo)/√

3)).

0

12

04

4 7 β010

1.5

B(q−p)m tcr

Fig. 6.5. Dependence of ultimate time tcr on βo and m for thick-walled elements

6.1 Use of Hoff’s Method 165

In a similar way the fracture of a cylinder can be considered. Using theprocedure above for the disk and the sphere we find

σθ − σr = 2µ(q − p)(b/r)2µ/(β2µ − 1), (6.8)

B(q − p)mt∞ = (4/3)(m/2)mβo∫

1

((β2 − 1)m/β(β2 − 1))dβ. (6.9)

For m = 1 and m = 2 we compute respectively

B(q − p)t∞ = (2/3) ln βo,B(q − p)2t∞ = (4/3) ln((βo + 1)2/4βo).

The consequent curves are drawn in Fig. 6.5 by broken lines 1, 2 and we cansee that for the tube t∞ is less than that for the sphere.

For a cone we use the condition of its constant volume (4.101) and intro-duce ratio

β = cos ψ/ cos λ.

Then λ is function of β as

cos λ = ((βo − 1)/(β − 1)) cos λo

and integral J6 in (5.146) is also a function of β. Now we find

dβ/dt = (β − 1)(tan λ)dλ/dt (6.10)

and sinceε(λ) = ln(sin λ/ sin λo)

(Fig. 3.24) thendλ/dt = tan λdε(λ)/dt

and from (5.141) with dε/dt, B instead of ε,Ω and (5.145) we derive

dε(λ)/dt = (3B/4)(q − p)m(cos λ)/(J6)m sin2 λ. (6.11)

Putting dλ/dt together with (6.11) into (6.10), separating the variables andintegrating as before we have finally

B(q − p)mt∞ = (4/3)(βo − 1) cos λo

1∫

βo

(J6)m(β − 1)−2dβ. (6.12)

The integrals in (6.12) should be calculated as a rule approximately.


6.1.4 Final Notes

Although the method in this sub-chapter uses somewhat unrealistic supposi-tion of an infinite elongation at the rupture, sometimes the fracture time isnear to test data. An analysis shows that the reason of it lays in the non-linearity of equations linking the rate of strains with stresses. Because of thatthe approach is widely used for the prediction of the failure moment of struc-tures. For example in /17/ a row of elements are considered. Among them agrating of two bars, thin-walled sphere and tube under internal pressure, along membrane loaded by hydrostatic pressure. Sometimes an initial plasticdeformation is also taken into account. The task of axisymmetric thin-walledshells under the internal pressure is formulated. An attempt of considerationof stress change on the base of creep hypotheses is also made. But the methodis mainly applied to a steady creep (see also Appendix L).

6.2 Mixed Fracture at Unsteady Creep

6.2.1 Tension Under Hydrostatic Pressure

For the bar in tension we use the notation of Fig. 6.3. According to relations(1.42), (1.45) we can link conditional qo and true q stresses by expression

q = qoe(1+α)ε

and from (2.32) we have

ε = 0.5Ω(t)(qo)m(e(1+α)ε + ko)m. (6.13)

By criterion dε/dt → ∞ we find from (6.13)

ε∗ = µ(κo exp(−(1 + α)ε∗) + 1)/(1 + α).

We can get critical time t∗ by putting ε∗ into (6.13). As we can see fromFig. 6.6 the critical strains (at α = 0) increase with a growth of the hydrostaticcomponent.

In the same manner the failure of the plate in the axisymmetric tensionunder hydrostatic pressure (Fig. 6.1) can be studied. As a result we have innotation of Sect. 6.1.1

(po)mΩ(t∗) = ε∗(exp(1 + α)ε∗ + κo)−m

and as we can see from Fig. 6.2 where for a steady creep (Ω = Bt) by solidlines at the same κo as the Hoff’s method diagrams t∗(µ) are constructedcritical time increases (similar to fracture time) with a fall of the hydrosta-tic component. We can also notice that t∗ < t∞ and it can be shown (seeAppendix L) that with a growth of creep curves non-linearity the differencebetween critical and fracture times increases. It can be explained first of all bythe circumstance that the method of infinite strain rate takes more realisticcondition of failure at finite strains than the Hoff’s approach.

6.2 Mixed Fracture at Unsteady Creep 167

0.5

1

ε*

0.5

0

1

2

µ0

Fig. 6.6. Dependence of ε∗ on µ at different κo for bar in tension under hydrostaticpressure

6.2.2 Axisymmetric Tension of Variable Thickness Plate with Hole

General Case and that of Constant Thickness

Here /29/ we use the same suppositions as in Sect. 6.1.2 which bring equation(2.33) to form

εe−αε = (3/4)Ω(t)(σθ)m (6.14)

where ε = εθ = ln(r/ro). Using the condition of a constant volume as

(ro + u)h = roho

with u = u(t) we integrate differential equation (5.130) in initial variables asfollows

(3Ω/4)µhobop =

b0∫

a0

ho(ro)(1 + u/ro)−1−αµ lnµ(1 + u/ro)dro. (6.15)

If we seek the critical state with the help of a computer we can applycriterion dε/dt → ∞ directly to expression (6.15). Calculated in this waydiagrams εb∗(β) (here β = bo/ao),ho = constant, α = 0 and α = 1 arerepresented in Fig. 6.7 by solid and broken lines 1. Critical time t∗ for thesecases at Ω = Bt is given from (6.15) in Fig. 6.4 by solid curves 2 and 1.

Curved Profile

For the case ho = hbbo/ro, α = 0 we derive from (6.15) at du/dt → ∞ equality

b0∫

a0

(1 + u/ro)−2(µ lnµ−1(1 + u/ro) − lnµ(1 + u/ro))(ro)−2dro = 0


0.8

ε*

0.4

01 4 7

32

42

1

0

β

- - -

Fig. 6.7. Dependence of ε∗ on β for thick-walled structures

or after computing the integral

(1 + ξ∗/β) lnµ(1 + ξ∗) = (1 + ξ∗) lnµ(1 + ξ∗/β)

where ξ = u/ao. This equation can be solved parametrically if we suppose

1 + ξ∗/β = (1 + ξ∗)η.

Here η is a parameter. We have from this equation

ξ∗ = ηµ/(η−1) − 1

and we can find other variables in form

β = ξ∗/((1 + ξ∗)η − 1), εb∗ = η ln(1 + ξ∗).

Curves εb∗(β) are given by dotted lines 1 and 2 in Fig. 6.7 for m = 1 andm = 2 respectively. When ξ∗ is known we can find t∗ from (6.15) as

pmΩ(t∗) = (4/3)

( β∫

1

(ξ∗ + ρ)−1 lnµ(1 + ξ∗/ρ)dρ

)m

where ρ = u/ao. Diagrams t∗(β) are drawn by dotted lines 1 and 2 in Fig. 6.4for m = 1 and m = 2 respectively. We can see that in this case the curvedprofile also gives higher critical time than that with ho = constant. Thisindicates that an optimal profile can be searched.

Optimal Profile

We shall seek such a disk among ones with radial cross-sections as following

ho = ao(β − 1)−1((ha − βhb)bo/(ro)2 + (β2hb − ha)/ro).


Putting this expression into (6.15) at α = 0 we receive

(3Ω/4)hbβp = ((β2hb − ha)

b0∫

a0

(1 + u/ro)−1 lnµ(1 + u/ro)dro/ro

+(bo(ha − βhb)(ln1+µ(1 + u/ao) − ln1+µ(1 + u/bo))/u(1 + µ))/(β − 1).

Using criterion du/dt → ∞ and integrating we find equation

(1 + µ)(1 − 1/β)ξ∗(ha(1 + ξ∗)−1 lnµ(1 + ξ∗) − βhb(1 + ξ∗/β) lnµ(1 + ξ∗/β))

= (ha − βhb)(ln1+µ(1 + ξ∗/β) − ln1+µ(1 + ξ∗)).

from which ratio ha/hb can be computed and we consider limit case ha = 0.For it we have at Ω = Bt

ln1+µ(1 + ξ∗) − ln1+µ(1 + ξ∗/β) = (1 + µ)ξ∗(1 + ξ∗/β)−1 lnµ(1 + ξ∗/β),

Bpmt∗ = 4(β∫1

(ρ + ξ∗)−1 lnµ(1 + ξ∗/ρ)dρ + (ln1+µ(1 + ξ∗/β)

− ln1+µ(1 + ξ∗))/(1 + µ)ξ∗)m/3(1 − 1/β)m.

Diagrams εb∗(β) for m = 1 and m = 2 are drawn in Fig. 6.7 as interruptedby points lines 1 and 2. We can see that they are somewhat higher than theconsequent values for profile ho = hbbo/ro. The curves t∗(β) for m = 1 andm = 2 are shown also by the same lines 1 and 2 as in Fig. 6.4 and they arelower than respective ones for curved profile above. However all the curvesfor every m are near to each other and from practical point of view the diskho = constant should be recommended.

6.2.3 Thick-Walled Elements Under Internaland External Pressures

Sphere

For this structure the initial equations are

εχ = εθ ≡ ε = ln(ρ/ρo) ≡ (1/3) ln(ρ/ρo)3,

ρ3 − a3 =(ρo)3 − (ao)3, εe−αε = Ω(t)(σθ − σρ)m.

(6.16)

Here ρ, χ = θ are spherical coordinates. The second of expressions (6.16) isthe condition of constant volume and using it for the whole sphere (ρ = b)as well as the first expression (6.16) we represent the strains at internal andexternal surfaces as

εa = (1/3) ln κ, εb = (1/3) ln κb (6.17)


whereκ = (a/ao)3, κb = (κ − 1)/(βo)

3, βo = bo/ao.

In order to integrate the static equation (2.80) in current variables wewrite the second (6.16) as following

1 − (ρo/ρ)3 = 1 − e−3ε = (a3 − (ao)3)/ρ3

and differentiate it. That gives

dρ/ρ = (1 − exp 3ε)−1dε.

Putting the last expression together with the third (6.16) into the first (2.80)we determine after integration

(q − p)(Ω(t))µ = 2

εa(κ)∫

εb(κ)

(1 − exp 3ε)−1(e−αεε)µdε. (6.18)

According to criterion dε/dt → ∞ we receive from (6.18)

κ−(1+αµ/3)∗ ln κ∗ = κb∗

−(1+αµ/3) ln κb∗

and if we suppose κ∗ = κb∗η the results can be represented as functions of

parameter ηκb∗ = η1/(η−1)(1+αµ/3),

after that we find κ∗ and

βo = ((κ∗ − 1)/(κb∗ − 1))1/3.

Considering in (6.18) the critical values of strains we get on the critical time.From Fig. 6.7 where for m = 1, α = 0 and α = 0.9 diagrams εb∗(βo) arerepresented by solid and broken lines 3 respectively we can see that the criticalstrains decrease with a growth of α and βo. In certain conditions (big α, βo, m)the strains at fracture can be small enough and it explains brittle destructionof structures made of plastic materials.

Comparing solid curve 3 in Fig. 6.7 with straight line 0 that correspondsto simple (see (1.46) at α = 0,m = 1) or biaxial equal tension we can see thateven thin-walled (βo is near to unity) sphere failures at strains 3 times lessthan a plate in the same stress state. From Fig. 6.5 where function t∗(βo) isshown by solid line 0 for the case of sphere at α = 0,m = 1 we can see thatt∗ is less than t∞.


Cylinder and Cone

In the same manner we consider the failure of a thick-walled tube /30/. Theresults can be represented in a form similar to (6.18) as

(q − p)mΩ(t∗) = (4/3)

( ε(κ∗)∫

ε(κb∗)

(e2ε − 1)−1(εe−αε)µdε

)m

(6.19)

wherein ε = εθ, κ = (a/ao)2, κb = 1 + (κ− 1)/(βo)2 and according to criterion

dε/dt → ∞ we haveκb∗ lnµ κ∗ = κ∗ lnµ κb∗.

A solution of the latter equation with a help of parameter η as in the cases ofthe disk and the sphere is also possible. Diagrams ε∗(β) and t∗(β) are drawnin Figs. 6.7 and 6.5 by solid lines 4.

A comparison of the results for the sphere and the cylinder allows toconclude that the fracture of the first one demands longer time than that ofthe second. However the strains at external surface in unstable state of thesphere are less than those of the cylinder and the latter are smaller for theplate in biaxial equal tension. All that can be explained by the influence ofthe form of a structure that is not taken into account by classical approachesto finding the ultimate state by the strength hypotheses.

In order to use the criterion of infinite strains rate to a cone we write thecondition of the constant material’s volume (4.101) in form

cos χ − cos χo = (δ − 1) cos ψo

where δ = cos ψ/ cos ψo. From these expressions and the definition of tangen-tial strain (Fig. 3.24) as

ε(χ) = ln(sin χ/ sin χo)

we derive equation

(1 − e−2ε) sin2 χ + 2(δ − 1) cos ψo cos χ − (δ − 1)2 cos2 ψo = 0

that allows to determine cos χ by solution

cos χ = (1 − e−2ε)−1(δ − 1) cos ψo +√

1 + (1 − e−2ε)−2e−2ε(δ − 1)2 cos2 ψ

and dχ is proportional to dε. With consideration of the last dependence andthe constitutive law we integrate (3.116) as follows

(3Ω/4)µ(q − p) =

εa(δ)∫

εb(δ)

εµe−(2+αµ)((δ − 1) cos ψo − (1 − e−2ε) cos χ)−1 cos χdε.

Applying to this relation criterion dε/dt → ∞ we get on an equation forcritical value of δ consideration of which gives the critical time. If its influenceis negligible we have a critical difference of pressures as before /31/.


6.2.4 Deformation and Fracture of Thin-Walled ShellsUnder Internal Pressure

General Relations

In the following we are going to consider a shell /32/ with initial length 2lobetween rigid bottoms with coaxial holes of radius a (in Fig. 6.8 a quarter ofthe structure is shown), diameter 2ro, the wall thickness ho under axial forceP and internal pressure q. We write down basic expressions for the shell’sdeformed state (solid lines in the figure). Static laws are:

σx = (q(r2 − a2) + P/π)/2rh cos θ,d(hrσx)/dr = hσy (6.20)

where σx, σy are stresses in longitudinal and circumferential directions,θ – angle between the shell’s ξ and x axes; relations for finite strains

εx = ln(dr/ sin θdξ), εy = ln(r/ro), εz = ln(h/ho) (6.21)

which are linked by constant volume condition

εx + εy + εz = 0.

The system is closed by a rheological expression which we take with aconsideration of the stress state at P = 0 as εx = 0. We suppose the secondconstitutive equation in form like (1.45) and (2.30) (here α is taken insteadof αµ)

2τe = ω(t)(γm)µ exp(−αε1). (6.22)

Equations (6.21) can be replaced by expression

νo =

εm∫

0

f(ε1)(sin θ)−1dε1. (6.23)

P

h

z

IIo

qr

r1

xθ

ξ

Rx

a

ro

Fig. 6.8. Thin-walled shell


13

4

4

0.8

0.4

0 2 4 νo

/ε*/

2 1'

Fig. 6.9. Dependence of critical strain on relative length of shell

Here f = exp ε1, νo = lo/ro, ε1 = εy = ε, εm = ε at ξ = 0 and function θ(εm, ε)can be derived from (6.20), (6.21) in form

(e2ε − χo)/ cos θ = exp 2εm − χo + Nq

ε∫

εm

εµe−αεdε (6.24)

where χo = (a/ro)2,Nq = 21+µω(t)ho/qro.With consideration of (6.24) we apply criterion dεm/dt → ∞ to (6.23). The

consequent diagram ε∗(νo) for m = 1, α = χo = 0 is given by solid line 1 inFig. 6.9. Computations show that at χo = 1 (the case of rings) the consequentcurve is near to that line. Critical time t∗ can be found from expression forNq at ε = ε∗.

Some Approximate Solutions

We demonstrate one of them for the case χo = 0,m = 1 when we have atεm − ε = η (η is a small value)

1/ cos θ = 1 + η(2 + Nq exp(−2εm)(exp(−αεm) − 1)/α),

νo =√

2εm exp(εm)(2 + Nq exp(−2εm)(exp(−αεm) − 1)/α)−1/2.

Using the criterion dεm/dt → ∞ we compute

Nq∗ = (2α(1 + 2ε∗) exp(2ε∗))/((1 + 4ε∗)(1 − exp(−αε∗)) − ε∗ exp(−αε∗)),

νo = exp ε∗(√

(1 + 4ε∗)(1 − exp(−αε∗)) − ε∗ exp(−αε∗)/√

2(1 − exp(−αε∗)) − α exp(−αε∗).

Constructed by broken line 1 in Fig. 6.9 according to the last expression curveis near to the rigorous one.


In the same manner we consider the case α = χo = 0 when we have

1/ cos θ = 1 + 2η(1 − Nq(εm)µ exp(−2εm)),

νo =√

εm(exp εm)(1 − Nq(εm)µ exp(−2εm)/2(1 + µ))−1/2.

By applying criterion dεm/dt → ∞ to the latter equation we find the criticalvalues

Nq∗ = (µ + 1)(1 + 2ε∗)/(ε∗)µ(1 − µ + 4ε∗),

νo =√

ε∗(exp ε∗)√

1 − µ + 4ε∗/√

2ε∗ − 1.

At m = 1 we have the results which follow from the previous approximatesolution for α = 0. If µ = 0.5 the consequent diagram is constructed inFig. 6.9 by solid line 2 and it is below the respected curve for m = 1.

Another Approximate Approach

In some works (/32, 33/ and others) the hypothesis is taken that straightlines in x directions deform as circular arcs with radius Rx (Fig. 6.8) definedby expression

Rx = 0.5((r1 − ro)2 + l2)/(r1 − ro).

That supposition fully determines the geometry and the deformation of theshell. Particularly

cos θ = (Rx − r1 + r)/Rx,

ε1 = ln(1 + 2ρ sin2(νo/2ρ))

where ρ = Rx/ro. Computing from (6.20), (6.21) σy and putting it togetherwith ε1 into (6.22) we find expression for ρ as follows

Nq2ρ lnµ(1 + 2ρ sin2(νo/2ρ)) = (2ρ cos2(νo/2ρ) − 1)(1 + 2ρ sin 2(νo/2ρ))2+α.

Using criterion dρ/dt → ∞ we derive an equation for ρ∗ according to whichinterrupted by points line1 is given in Fig. 6.9. The dotted curve is constructedin the supposition of l = lo /33/. The good agreement of these results with therigorous solution (especially at ν > 5) opens the way for approximate studyof other shells.

Torus of Revolution

We suppose that the structure (one quarter of which is shown in Fig. 6.10)changes at deformation its dimensions but not the form. The measurementsand natural observations show that

Ro − ro = R − r


R

q

h

y

rz

σy

σy

Fig. 6.10. Torus of revolution

and the fracture takes place in points y = 0, z = r. So, for them at h =constantwe have

Rhr = Rohoro(this expression is valid also for the whole volume of the material atho= constant). The stresses there are

σy = σoe2ε(ρo + eε)/(ρo + 1)

where σo = qro/ho, ρo = Ro/ro − 1, 0 < ρo < ∞ and strain ε = ε1 = εy isdetermined by the second expression (6.21).

Deformation εz is given by the third relation under this number and forεx we write

εx = ln(R/Ro),

So, with consideration of constant volume condition we find maximum shear

γm = 2ε + ln((ρo + eε)/(ρo + 1))

and rheological law (6.22) becomes

σoe(2+α)ε(ρo + eε) = ω(t)(2ε + ln((ρo + eε)/(ρo + 1))µ(ρo + 1).

According to criterion dε/dt → ∞ we derive equation for critical deformationas

((2+α)(ρo+exp ε∗)+exp ε∗)(2ε∗+ln(ρo+exp ε∗))/(ρo+1) = µ(2ρo+3 exp ε∗).

At ρo → ∞ (a long tube) and ρo = 0 (a sphere) we compute respectivelythe relations that were received earlier for these structures

ε∗ = 1/(2m + α), ε∗ = 1/(3m + α)

and that confirms a validity of the hypotheses above. The consequent curvesfor α = 0 are drawn in Fig. 6.11 by solid straight lines t, s and we can expectthat other cases are between them.


0.6

0.4

25

36

4 71

0.2

0 0.25 0.5

s

t

0.75 µ

ε*

Fig. 6.11. Dependence of critical deformations on µ

Oo

Ro

y

LLoA

BR

O

θo

θβ

βo

Ao

Fig. 6.12. Cross-section of membrane

6.2.5 Thin-Walled Membranes Under Hydrostatic Pressure

General Expressions and Cylindrical Membrane

These structures are more often used in the Geo-mechanics (see e.g./34/).We begin with a long membrane as a part of a cylinder the cross-section ofwhich is shown in Fig. 6.12 and suppose as before that its form remains ata deformation (solid line in the figure). From the condition of the constantmaterial’s volume and geometrical considerations we have

Lh = Loho,Lo = Roθo,L = Rθ,Ro sin θo = R sin θ (6.25)

where h is a thickness of the membrane. We shall solve the problem in termsof angle θ when for ε ≡ εy and σy = 2τe we receive with considerationof (6.25)

ε = ln(θ sin θo/θo sin θ), σy = σoθ sin2 θo/θo sin2 θ.


Here σo = qRo/ho and q is the hydrostatic pressure. Putting ε and σy into(6.22) we deduce

θ sin2 θo/θo sin2 θ = Nqα ln(θ sin θo/θo sin θ)

where Nq has the same value as in (6.24) and α is taken equal to zero.Using criterion dθ/dt → ∞ we find equation for critical θ

µ = ((sin θ∗ − 2θ∗ cos θ∗)/(sin θ∗ − θ∗ cos θ∗) + α) ln(θ∗ sin θo/θo sin θ∗)

which at θo = θ∗ = π (a cylinder) gives ε∗ = µ/2 as for a long tube (solidstraight line t in Fig. 6.11) and for another ultimate case θo = 0 we have thesolid curve. So we can conclude that at 0 < θ < π the points are betweenthese lines and the influence of a form of the structure on its ultimate stateis high.

Spherical Membrane

For such a segment we keep picture 6.12 and take condition

ε ≡ εx = εy = −εz/2.

Then with consideration of (6.25) write for stresses

σ ≡ σy = σx = qRoe2ε(sin θo/ sin θ)/2ho.

Putting σ and ε into (6.22) we derive

qRo(sin θo/ sin θ)/ho = 3µ2εµe−(2+α)εω(t).

According to criterion dε/dt → ∞ and after an exclusion of ω(t) togetherwith the constants we find

µ = (2 + α − (dθ/dε)∗ cot θ∗)ε∗. (6.26)

Further we can take different suppositions. The simplest of them ish = constant, but strains can be calculated in two options. From the con-dition of material incompressibility we have

ε = 0.5 ln((1 + cos θo)/(1 + cos θ))

and from (6.26)

µ = (0.5α + (1 − 2 cos θ∗)/(1 − cos θ∗)) ln((1 + cos θo)/(1 + cos θ∗)).

If we compute ε according to a change of a meridian then similar to the caseof the cylindrical membrane we receive

µ = (α + (2 tan θ∗ − 3θ∗)/(tan θ∗ − θ∗)) ln(θ∗ sin θo/θo sin θ∗). (6.27)


The calculations show that both options give similar results that for the caseα = θo = 0 are constructed by broken curve in Fig. 6.11. Its comparison tostraight line s for the sphere shows also the big influence of the form on thecritical state of this structure..

The case of non-homogeneous deformation (when h is not a constant) weconsider on the base of the supposition that point Ao on straight line OoB inFig. 6.12 comes to point A of ray OB. From the figure we find the expression(valid at θo < π/2) for ε in the pole where β = βo = 0

ε = − ln(cos θ/ cos θo)

putting which into (6.26) we receive

µ = −(2 + α − cos2 θ∗) ln(cos θ∗/ cos θo). (6.28)

The consequent curve is drawn in Fig. 6.11 by interrupted by points line atα = 0.

Comparison with Test Data

In Fig. 6.12 the experimental points are given according to /35/ for mild cop-per (1 – Brown–Sacks), aluminum, hard and mild steel (2, 3, 4 – Brown-Tomson), two types of copper (5, 6 – Weil-Newmark), polyethylene (7 from/36/). We can see that these points are situated between curves according to(6.27), (6.28). As the first of them gives safer values it can be recommendedfor practical use. We can also note that solution (6.27) at µ = 1, α = θo = 0gives θ∗ near to represented in /37/ where deformation ε = θ/ sin θ and linearlink between σ, ε are taken. This θ∗ corresponds well enough to test data onbutadiene rubber membranes of different thickness. It opens the way to thetheory above also for rubbers.

So, the test data above confirm the theory in the case of a spherical seg-ment. The good agreement of this approach was fulfilled by the author forthin-walled tubes under axial load, internal pressure and torsion (see /14, 38/and others).

6.2.6 Two Other Problems

Tension of Tube of Limited Length

Now we consider the case of axial load P action when we suppose that indangerous cross- section ξ = 0 the strains are linked as

εy = εz = −εx/2

(a hypothesis of the material). We rewrite the second equilibrium expression(6.20) with consideration of (6.21) as follows


2 exp εxd(exp(−εx)σx)dεx = −σy

or taking into account (6.22)

d(σx exp(−εx/2)) = 0.5(3/2)µω(t) exp(−(α + 0.5)εx)dεx.

An integration of this expression together with (6.21) and the first (6.20) givesat εx ≡ ε

1/ cos θ = e−ε/2(exp(εm/2) + 0.5NP

ε∫

εm

εµe−(α+0.5)εdε) (6.29)

whereNP = (3/2)µω(t)σo, σo = P/2πroho.

Now we apply criterion dε/dt → ∞ to (6.23) with f(ε1) = exp(−3ε/2). Com-puted with a help of a PC curve for α = 0, m = 1 is given in Fig. 6.9 bysolid line 3 in which broken straight line corresponds to the bar of infinitelength. We can see that the influence of the latter takes place only for smallνo. Critical time t∗ can be found from the expression for NP at εm = ε∗.

Compression of Cylinder

The solution above can be used for a compression by rough plates of a shortcylinder (see Fig. 4.38) if we suppose that it consists of a set of thin-walledtubes. Computed with a help of PC diagram is represented by solid line 4in Fig. 6.9. The difficulties of this task solution and its importance allow toapply some approximate approaches similar to that for the thin-walled shellunder internal pressure above.

We rewrite approximately (6.29) as

1/ cos θ = 1 + 0.5η + 0.5NP(0.5 + α)−1(η exp(−(1 + α)εm)+ exp(−εm/2)(exp(−(0.5 + α)εm) − exp(−(0.5 + α)ε))/(0.5 + α)

where η = εm−ε. Decomposing in the last expression and in (6.23) exponentsin series we have for α = 0, m = 1

1/ cos θ = 1 + 0.5η(1 − NPεm),

νo = (exp(−3εm/2))√

εm/√

1 − NPεm.

Using criterion dε/dt → ∞ we finally find

NP∗ = (3ε∗ − 1)/3(ε∗)2,νo =

√3/ε∗ exp(−3ε∗/2)

The diagram of the latter expression is given in Fig. 6.9 by broken line 4and we can see that it is near to the solid line under the same index at nothigh νo.


Final Notes

The relations above are important because the theory supposes infinite lengthof samples although the experiments are usually made on the ones of limitedlength and so we can appreciate its influence on the results.

6.2.7 Ultimate State of Anisotropic Plate in Biaxial Tension

General Considerations

The application of this sub-chapter method to a plate in a tension allowsnot only to find the ultimate state of similar structure elements but also toestablish a theoretical strength of an element of a material. The latter problemis usually solved in an experimental way with a help of so-called strengthhypotheses.

Basic Expressions

To receive initial equations we use the links between true σs and conditionalσso stresses in form (1.41) which result from constant density condition. Byapplying these relations to initial rheological laws (2.39), (2.43) we obtain theexpressions describing a development of strains in time as follows

εxSy = εySx, exp(−αεeq)εx = Ω(t)(σxo)mDm−1Sx (6.30)

where D = σeq/σxo and the values of Sx, Sy depend on the position of thesymmetry axis relatively to the loading plane. For the cases when z, x, y areisotropy axes we have

Sx = exp εx − kno exp εy,Sy = no exp εy − kexpεx,

Sx = (1 − k)(2 exp εx − no exp εy),Sy = no exp εy + (k − 1) exp εx,

Sx = exp εx + (k − 1)no exp εy,Sy = (1 − k)(2no exp εy − exp εx).

Here no = σyo/σxo.

Ultimate State

Any stage of the deformation can be taken as a limiting one but the mostconvenient is the use of criterion dε/dt → ∞ when at no = constant we findfrom (6.30) an equation for critical strains as follows

(ε∗)2(1 + R∗)no(1 − k2) exp(ε∗ + εy∗) − C∗Sx∗ε∗ + (Sx∗)2 = 0. (6.31)

Here ε∗ = εx∗, R = (m − 1)∂D/∂εx + α and functions C(no, εx, εy) for theoptions above are respectively


C = RSx + exp εx + no exp εy,

C = RSx + 2(1 − k) exp εx + no exp εy,

C = RSx + exp εx + 2(1 − k)no exp εy.

(6.32)

From Figs. 6.13 and 6.14 where for the cases when z and y are axes of symmetry(the case if x is axis of isotropy is similar of the first of them – this similaritywas explained in Sect. 2.3.3) diagrams mε∗(m, k,no) for α = 0, m = 1 andm → ∞ (signs 1 and b) are represented we can see the great influence oncritical state of the structure of m and k (solid, broken, interrupted by pointsand dotted lines for k-values equal to 0.5 (isotropic material – see Sect. 2.3.3),1, 0,−1 respectively) and sign A means a validity for both m above.

Ultimate State of Plastic Materials

Expression (6.31) does not contain time and hence it can be used for plasticmaterials the limit state of which is usually evaluated by the dependence of

A

b

0.75 no0.50.250

0.5

mε*

1

Fig. 6.13. Dependence of mε∗ on m, k and no when z is axis of symmetry

1

1

b

0.75 no0.50

0.5

mε*

0.25

Fig. 6.14. Dependence of mε∗ on m, k and no when y is axis of symmetry


critical sub-tangent Z∗ /39/ (see Fig. 2.6)

Z∗ = σe∗(dεe/dσe)∗

to curve σe(εe) on n. The transition to these variables can be made as

Z = mεe,no = nexp(εx − εy) (6.33)

and after transformations we obtain

(Z∗√

3/2√

Λ∗)2n∗(1 − k2)(1 + α/m) − (Z∗√

3/2√

Λ∗)C1∗ + m = 0

where Λ are given by (2.40), (2.44) and values of C1(n) can be derived from(6.32) as follows

C1 = (m − 1 + α)(1 − kn) + 1 + n,

C1 = (m − 1 + α)(1 − k)(2 − n) + 2(1 − k)n,

C1 = (m − 1 + α)(1 − (1 − k)n) + 1 + 2(1 − k)n.

(6.34)

The analysis shows (see Figs. 6.15 and 6.16 in which diagrams Z∗(n) are con-structed for the same options and in the same manner as in Figs. 6.13 and6.14) that parameters k, n, m have high influence on the ultimate plasticityof the body. Sometimes the same material can reveal absolute rigidity as wellas infinite strains. We must notice that at m → ∞, α = 0 we have relationfor Z∗ in “classical” /39/ sense as

Z∗ = 2√

Λ/√

3c∗where for the options above

c = 1 − kn, c = 1 + (k − 1)n, c = (1 − k)(2 − n).

For isotropic material at k = 0.5 and σeq = 2τe we receive

Z∗ = 2√

1 − n∗ + (n∗)2/(2 − n∗).

A

b

1

0.50

1

Z*

n

Fig. 6.15. Dependence Z∗(n) when z is axis of symmetry


A

b

1

1

Z*

0 0.5 n

Fig. 6.16. Dependence Z∗(n) when y is axis of symmetry

Ultimate State for Brittle Materials

When α is big we can neglect m−1 and m in expressions (6.34) and for R∗.So, putting the second relation (6.33) into (6.31) we receive

ε∗2n(1 − k2) − ε∗c2 + c2 = 0. (6.35)

Solving this equation and decomposing the function under square root in serieswe have for two members of it a very simple and important result

ε∗ = 1/α (6.36)

which coincides with the first relations (1.47), (5.129), (5.134) and others ofthat kind for all the structures in any stress state and can be considered as a“law of nature”. Putting (6.36) into the second expression (6.30) we find thecritical time

Ω(t∗) = 1/αeσxomD∗

m−1Sx∗.

When the influence of time is negligible we deduce for Ω = constant

σxo∗ = (ΩαeD∗m−1Sx∗)µ.

In the same manner the critical state of anisotropic thin-walled tubeunder axial load and internal pressure in three options of symmetry axis andloading plane as well as with addition of torsion (for case k = 0, 5) can bestudied /40/.

Conclusion

In the book above we tried to put the bases for computations in theGeo-mechanics as a whole and the Soil Mechanics in particular forthree types of body: elastic, plastic and hardening at creep with a dam-age. We demonstrated the methods of consequent sciences on the mostimportant for practice examples (a slope under vertical loads, a compressedwedge, an action of a soil on a retaining wall, a flow of a material between twofoundations, inclined plates and in a cone, a wedge and a cone penetration aswell as the load-bearing capacity of piles and sheets of them, a tension andtransversal shear of a body with a crack and a pressure on it of a punch, ulti-mate state of sphere, cylinder and cone under internal and external pressures,and plates at bending etc). Some of the problems were not studied due to themathematical difficulties (e.g. a slope under combined loads for a non-linearmaterial) or thanks to their unclear mechanical formulation.

We suppose as the most important Chaps. 5 and 6 that describe processesin structures between an initial elastic deformation and the final ultimatestate, representing them as particular cases. To the regret Chap. 5 is the mostdifficult from mathematical point of view. However proposed here method ofreducing the third order differential equations to the systems of the first orderones that can be solved one after another simplifies the procedure. We hopethat this approach can be useful for some other tasks. We give also wheneveris possible simple engineering relations.

The solutions in Chap. 6 take into account large displacements and strainsthat are often met in the geo-mechanical and other processes. Here the appli-cation of the criteria of infinite elongations and their rates is very convenient.The phenomenon of unstable change of strains due to a damage is also oftenmet. The experiments show that the prediction of the method is nearer toa reality for more unsteady creep and non-linear link between stresses andstrains. If we suppose that in the common sense the destruction is obliged tomain elongations in dangerous body parts the method is widely used. Thanksto consideration of a damage (it is well-known that the theoretical strengthof a material is much higher than a real one) this approach can be applied

186 Conclusion

to structures that are destroyed at small change of dimensions and it wasdemonstrated in Chaps. 5 and 6. Here we underline as the most importantresult the independence of the maximum critical strain in brittle materials onthe types of structure and stress state.

In appendixes some auxiliary data are given. Among them the tablesfor simple computations of necessary values, details on calculations of somecomplex relations, a study of the fracture of brittle materials at eccentric com-pression, bases of applied creep theory similar to that in plasticity, the use ofcreep hypotheses for determination of the fracture time and others.

Most results of the book have been received by the author in recent yearsand they are published in different editions throughout the world(additionallyto the author references above see also /41/. . ./47/). We hope that their com-pact presentation together with some well-known achievements in the field isnecessary.

Herewith I use the occasion to remember the professor of Leningrad StateUniversity L.M. Kachanov who opened me the ideas of new approaches toFracture Mechanics, and the professor of Odessa State University I.P. Zelinskywho put me for a solution some geo-technical problems. I also want to thankthe workers of CIP Insel of Electrotechnic faculty of Bochum University whohelped me to create electronic version of this book.

S. Elsoufiev, Bochum, BRD 2006.

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Noordhoff., 1975. 637p7. Elsoufiev S.A. Modern methods of strength computations. Odessa, 1990, 60p

(In Russian)8. Griffith A. The phenomenon of rupture and flow of solids. //Phil. Trans. of

Royal Society. 1920 A-221, 163. . . 1989. Gvozdev A.A. Development of the theory of reinforced concrete in the U.S.S.R.

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University. 1998, 256p (In Russian)11. Coulomb Ch. Essai sur une application des regle maximis et minimis a quelque

problemes de static relatifs a l’architecture. Mem. Acad. Royal Pres. DiversSavants. Paris, 1776, v.7

12. Timoshenko S.P. History of strength of materials. N.-Y. McGraw-Hill Book Co.1953, 225p

13. Hoff N. The necking and the rupture of rods subjected to constant tensile loads.//Journ. of Applied Mechanics, 1953, v.20, 105. . . 108

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15. Carlsson R. Creep induced tensile instability. //IJ of Mech. Sci. 1965, 7, 2,218. . . 222

16. Elsoufiev S.A. To the solution of some nonlinear problems in Geomechanics.Proc. of 10th Danube-European Conference on SM&EF. 1995, 553. . . 558

17. Kachanov L.M. Introduction to continuum damage mechanics. Dordrecht, 1986,280p

18. Sokolovski V. Theory of Plasticity. Moscow, 1969, 608p (In Russian)19. Panasjuk V. Fracture mechanics. Successes and problems. Collection of abstracts

of 8th IC on Fracture. Lviv (Ukraine), 1993, p.3

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of the Third IC on Ground Improvement Geo–systems. L-n, 3-5 June 1997,329. . . 334

24. Kachanov L.M. Foundations of the theory of plasticity. Amsterdam, 1971,482 p

25. Elsoufiev S.A. Ultimate state of a slope at non-linear unsteady creepand damage. In Proc. of IS on Slope Stability Engineering. A.A.Balkema/Rotterdam/Brookfield/, 1999, 213. . . 217

26. Kachanov L.M. The creep of a thin layer pressed by rigid plates. //Izvestija ofAN USSR. Department of Technical Sciences. 1965, 6 (In Russian)

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finding according to the method of transition to unstable condition. //Machi-novedenie, 1982, 5. 94-96 (In Russian)

33. Weil N. Tensile instability in thin-walled cylinders of finite length. //J. of Mech.Sci. 1963. V.5, 487–490

34. Study of waterproofing revetments for the upstream face of concrete dams. Finalreport. GIGB, LCOLD, 1998, 164p

35. Weil N. Rupture characteristics of safety diaphragms. //Trans. of Am. Soc. ForMech. Eng. (JAM), 1959, v.4, 621. . . 627

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37. Natov N., Koleva V. Investigation of polymer plates at biaxial stress state.//Mech. of composite materials. 1982, 3, 489. . . 492 (In Russian)

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39. Lankford W., Saibel E. Some problems of unstable plastic flow under biaxialtension. //Trans. of Amer. Institute of Mech. Engineers. Inst of Metal Division.1947, 171, 562. . . 571

40. Elsoufiev S.A. Strength of anisotropic tubes at different loading. In Proc.of the Sixth European Mechanics of Materials Conference. Non-linearMechanics of Anisotropic Mat, University of Liege/Belgium. Sept. 9–12 2002,227. . . 232

41. Elsoufiev S.A. On bearing capacity of piles and sheets of piles. In “Grouting SoilImprovement Geosystems Including Reinforcement.” Helsinki. 2000, 457. . . 465

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43. Elsoufiev S.A. Propagation of cracks and plastic zones at cycling loading. Ex-tended Abstracts to Symposium on fatigue testing and analysis under variableamplitude loadings condition. Tours (France). 29. . . 31 May 2002, 1. . . 3 (No 68)

44. Elsoufiev S.A. Fracture of shells and membranes at non-linear unsteady creepand damage. In Abstracts of 76th Annual Scientific Conference of Gesellschaftfur Angewandte Mathematik und Mechanik e.V. Part 2, 28 March-1 April 2005,146. . . 147

45. Elsoufiev S.A. Rheology and fracture of composite materials. In book ofAbstracts of IUTAM Symposium “Multiscale Modelling of Damage and Frac-ture Processes in Composite Materials.” Kazimerz Dolny, Poland, 23. . . 27 May2005, p.9

46. Elsoufiev S. Fracture due to cracks and plastic zones near stamp edges propa-gation. In Abstracts of EC 466 “Computational and Experimental Mechanicsof Advanced Materials.” 20–22 July 2005. Loughborough, UK, p.79

47. Elsoufiev S.A. Tarasova B. English in texts on the Strength of Materials. Odessa,1994, 92p

48. Il’jusin A.A. Plasticite.(Deformations elastico-plastique). Paris: Eyrolles. 195649. Butin V., Elsoufiev S.A. Estimation of resistance to deformation of locks

sliding supports. //Trudy LIWT No 158. Waterways and hydrotechnical struc-tures. Leningrad. 1977, 156. . . 163. (In Russian)

A

Appendix

Computation of p* for Brittle Materialswhich do not Resist Tension

In Germany the computation of ultimate load for brittle material is usuallymade according to the schemes that are represented in Fig. A.1.

If e = h/6 force F is in the edge of the core /47/ (Fig. A.1, a) and themaximum stress is

p∗ = 2F/h

and F is the first ultimate load

a)

b)

c)

h/2 h/2

e

e

e

F

p*

p*

p*

F

F

Fig. A.1. Distribution of stresses at eccentric in brittle materials like soils

192 A Appendix

If h/6<e<h/3 (the force is out of the core – Fig. A.1, b) we receive fromstatic equation

p∗ = 4F/3(h − 2e).

If e = h/3 (damage reaches the axis of symmetry – Fig. A.1, b) we have

p∗ = 4F/h

and F is the second ultimate load which in this case is less than the first one.

B

Appendix

Values of coefficients for computation of compressive stresses due to concentratedforce

r/z K r/z K r/z K r/z K r/z K r/z K

0.00 0.4775 0.32 0.3742 0.64 0.2024 0.96 0.0933 1.28 0.0422 1.60 0.02000.02 0.4770 0.34 0.2632 0.66 0.1934 0.98 0.0887 1.30 0.0402 1.62 0.01910.04 0.4756 0.36 0.3521 0.68 0.1846 1.00 0.0844 1.32 0.0383 1.64 0.01830.06 0.4732 0.38 0.3408 0.70 0.1762 1.02 0.0803 1.34 0.0365 1.66 0.01750.08 0.4699 0.40 0.3294 0.72 0.1684 1.04 0.0764 1.36 0.0348 1.68 0.01670.10 0.4657 0.42 0.3181 0.74 0.1603 1.06 0.0727 1.38 0.0332 1.70 0.01600.12 0.4607 0.44 0.3068 0.76 0.1527 1.08 0.0694 1.40 0.0317 1.72 0.01530.14 0.4548 0.46 0.2955 0.78 0.1455 1.10 0.0658 1.42 0.0302 1.74 0.01470.16 0.4482 0.48 0.2843 0.80 0.1398 1.12 0.0626 1.44 0.0288 1.76 0.01410.18 0.4409 0.50 0.2733 0.82 0.1320 1.14 0.0595 1.46 0.0275 1.78 0.01350.20 0.4329 0.52 0.2625 0.84 0.1267 1.16 0.0567 1.48 0.0263 1.80 0.01290.22 0.4242 0.54 0.2518 0.86 0.1196 1.18 0.0539 1.50 0.0251 1.82 0.01240.24 0.4151 0.56 0.2414 0.88 0.1138 1.20 0.0513 1.52 0.0340 1.84 0.01190.26 0.4054 0.58 0.2313 0.90 0.1083 1.22 0.0489 1.54 0.0229 1.86 0.01140.28 0.3951 0.60 0.2214 0.92 0.1031 1.24 0.0466 1.56 0.0219 1.88 0.01090.30 0.3849 0.62 0.2117 0.94 0.0981 1.26 0.0443 1.58 0.0209 1.90 0.0105

C

Appendix

Values of coefficients K for computation of stresses by the method of corner points

n= m = a/bz/b

1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 3.0 4.0 8.0

0.0 0.250 0.250 0.250 0.250 0.250 0.250 0.250 0.250 0.250 0.250 0.250 0.2500.2 0.249 0.249 0.249 0.249 0.249 0.249 0.249 0.249 0.249 0.249 0.249 0.2490.4 0.240 0.242 0.243 0.243 0.244 0.244 0.244 0.244 0.244 0.244 0.244 0.2440.6 0.223 0.228 0.230 0.232 0.232 0.232 0.233 0.233 0.234 0.234 0.234 0.2340.8 0.200 0.202 0.212 0.215 0.215 0.217 0.218 0.219 0.219 0.220 0.220 0.2201.0 0.175 0.185 0.191 0.196 0.198 0.200 0.201 0.202 0.203 0.203 0.204 0.2051.2 0.152 0.163 0.171 0.176 0.179 0.182 0.184 0.185 0.186 0.187 0.188 0.1891.4 0.131 0.142 0.153 0.157 0.161 0.164 0.167 0.169 0.170 0.171 0.173 0.1741.6 0.112 0.124 0.133 0.140 0.145 0.148 0.151 0.153 0.155 0.157 0.159 0.1601.8 0.100 0.108 0.117 0.124 0.129 0.133 0.137 0.139 0.141 0.143 0.146 0.1482.0 0.084 0.095 0.103 0.110 0.116 0.120 0.124 0.126 0.128 0.131 0.135 0.1373.0 0.045 0.052 0.058 0.064 0.069 0.073 0.078 0.080 0.083 0.087 0.093 0.0984.0 0.027 0.032 0.036 0.040 0.044 0.047 0.051 0.054 0.056 0.060 0.067 0.0755.0 0.018 0.021 0.024 0.027 0.030 0.033 0.036 0.038 0.040 0.044 0.050 0.0607.0 0.009 0.011 0.013 0.015 0.016 0.018 0.020 0.021 0.022 0.025 0.031 0.04110.0 0.005 0.006 0.007 0.007 0.008 0.009 0.010 0.011 0.012 0.013 0.017 0.026

D

Appendix

Data for construction of centre of the most dangerousslip arc when soil has only coherence

Gradientof slope

in degrees

β1 β2

1.73:1 29 401:1 28 371:1.5 26 351:2 25 351:3 25 351:5 25 37

E

Appendix

Values of coefficients A, B for approximate computation of slope stability

Gradientof slope1 : m

Slip surfacegoes throughlower edgeof slope

Slip surface goes through base of slopeand has horizontal tangent in depth

l = h/4 l = h/2 l = h l = 1.5h

A B A B A B A B A B1 : 1.00 2.34 5.79 2.56 6.10 3.17 5.92 4.32 5.80 5.78 5.751 : 1.25 2.64 6.05 2.66 6.32 3.24 6.62 4.43 5.86 5.86 5.801 : 1.50 2.64 6.50 2.80 6.53 3.32 6.13 5.54 5.93 5.94 5.851 : 1.75 2.87 6.58 2.93 6.72 3.41 6.26 4.66 6.00 6.02 5.901 : 2.00 3.23 6.70 3.10 6.87 3.53 6.40 4.78 6.08 6.10 5.951 : 2.25 3.18 7.27 3.26 7.23 3.63 6.56 4.90 6.16 6.18 5.981 : 2.50 3.53 7.30 3.46 7.62 3.82 6.74 5.08 6.26 6.26 6.021 : 2.75 3.59 8.02 3.68 8.00 4.02 6.95 5.17 6.36 6.34 6.051 : 3.00 3.59 8.91 3.93 8.40 4.24 7.20 5.31 6.47 6.44 6.09

F

Appendix

Computation of Stresses at Anti-Plane Deformationof Massif with Crack

We put in (5.6) C-θ = x and rewrite it as

f4 = D4((√− (m − 1) cos x)/(

√+ (m − 1) cos x))(m−1)/4(m+1)

× ((√

+ (m + 1) cos x)/(√− (m + 1) cos x)) sin2 x

where√

=√

(m + 1)2 − (m − 1)2 sin2 x.

At x = 0 we have a peculiarity of 0/0 type. Using the l’Hopital’s rule wereceive D. In the same manner we find that f ′(0) = 0. Then we use boundarycondition f(π) which gives C = 0.

Now we find according to the procedure above

f ′(π) = Dmm/2(m+1)/√

m + 1 ≡ D2.

Lastly we compute f, f′ at θ = π/2 as

f(π/2) = D, f ′(π/2) = D√

m/(m + 1)

Table F.1. Values R(θ) at different m

m 1 3 15θ0 0.318 0.512 0.786π /4 0.318 0.445 0.579π /2 0.318 0.295 0.1913π /4 0.318 0.201 0.070π 0.318 0.173 0.052

202 F Appendix

and rewrite (5.8) as following

R(θ) = (√

(mf/(m + 1))2 + f ′2)m+1/I(m)/f ′(π)/m

whereR = 2GΩ(t)(τe)m+1r/πτ2l.

The values of R are given in the table earlier and according to it Fig. 5.2is constructed.

G

Appendix

Some Computations on Bending of Wedge

Some materials (including soils) have stress-strain diagram with µ > 1. Weconsider the case µ = 2 when from (5.71) and boundary conditions we have

g cos 2ψ = Hsin θ.

Then from (5.70) we derive differential equation

dψ/dθ = sin(θ − 2ψ) cos 2ψ/(1 + sin2 2ψ) sin θ. (G.1)

We integrate (G.1) at different Θ(0) = Θo by the finite differences method.The diagrams ψ(θ) for Θo = 1, 2, 3, 4 are given in Fig. G.1 by curves 1, 2, 3,4 respectively. The consequent values of λ are 40, 64, 87 and 117.

Then according to (5.69) we have

τ = (ωH2/r2))(sin θ/ cos 2ψ)2 sin 2ψ

300

−15

−30

Ψ

θ860

2

1

3

4

90

Fig. G.1. Diagram ψ(θ) for µ = 2

204 G Appendix

and with consideration of (5.72) and (G.1) we compute

max τe = MΘo2/2Jr2

where

J =

λ∫

0

(sin θ/ cos 2ψ)2 sin 2ψdθ.

The diagram max τer2/M as a function of λ is given in Fig. 5.13 by interruptedby points line and we can see that it is much higher than that one for m = 1.

As particular case of the general theory we take it for m = 2 when (5.78)becomes

dΘ/dψ = −2Θ(1 − Θ)(1 + 2(1 − Θ)/Ψ) (G.2)

where Ψ = (1+cos2 2ψ)/2. We integrate this equation at Θo equal to 0.5 and1.249, and draw the curves ψ(θ) in Fig. G.2 with numbers 0.5 and 1.25 (forthe case Θo = Θ = 1 we have straight line 1 with relation ψ = θ − π/4). ForΘo > 1.25 we cannot find λ since at small ψθ → ∞ (e.g curve 1.4 for Θo = 1.4in the figure). Computing by (5.82) M-value we obtain max τe according to(5.81) which is shown by pointed line in Fig. 5.13, and we can see that it isnear the solid curve (for m = 1) and, so, for Θo > 1.25 the latter must beused as the first approach (similar to the case in Sect. 5.2.2).

In conclusion we derive similar to Sect. 5.2.2 an engineering solution. Wetake with consideration of (5.66) rheological law like (5.30)

εθ = −3Ω(t)r−2µF(m−1)/2f ′/4, γ = 3Ω(t)r−2µF(m−1)/2f

where F = f ′2+4f′2 and put it into (2.71) which leads to non-linear differential

equation

(m − 1)((m − 3)(f ′′ + 4f)2f′2 + F(f ′′′ + 4f ′)f

′2 + 3F(f ′′ + 4f ′)f ′f ′′ − 4mF2f ′)

+ f2f ′′′ + 4(2m − 1)((m − 1)F(f ′′ + 4f)ff ′ + F2f ′) = 0.

0 20

0.5 1

1.251.4

−10

−20

−30

Ψ8

θ840 60

Fig. G.2. Curves ψ(θ) for case µ = 0.5

G Appendix 205

At m = 1 it gives the solution of Sect. 5.2.5. If we suppose f ′ = 0 in thewhole wedge we derive simple expression f ′′′ = 0 which with consideration ofboundary conditions (5.72) and static demand (5.72) provides solution

f = 3M(θ2 − λ2)/4λ3.

The value of max τe takes place at θ = 0 and is

max τe = 3M/4r2λ. (G.3)

The computation show that at π/6 < λ < π/4 max τe is near to the curves forµ = 0, µ = 0.5, µ = 1 in Fig. 5.13 and at π/4 < λ < π/2 it is somewhat belowof the solid line (µ = 1). So, in these limits engineering relation (G.3) can beused as the first approach.

H

Appendix

Bases of Applied Creep Theory

As was shown in Sect. 4.2.2 a tension or a compression of a thin layer inducesat plane deformation almost equal triple-axial stress state with small shearingcomponents. It opens the way to the construction of the applied plasticitytheory /48/. In a similar manner the applied creep theory can be formulated.

We create it for the part I in Fig. 5.14. Since σo does not depend on y wewrite down the first equation (2.59) at X = 0 as

dσx/dx − 2τ/h

or after the use of the Coulomb’s law (4.44)

dσx/dx = −2fσy/h. (H.1)

As was shown in /49/ strain εx is a constant (its dependence on time is furtherhinted) and according to (5.84) σx, σy are functions of x only. Putting (5.84)into (H.1) and integrating we receive

σy = (σo − 4ω(t)(εx)µ)(e2f(b−a)/h − 1) (H.2)

Now we find the specific pressure in y direction as following

N/2a = (l/a)

a∫

0

σy(x)dx.

Replacing here σy from (H.2) we have after integration relation

N/2a = (σo − 4ω(t)(εx)µ)h(e2fa/h − 1)/2fa (H.3)

from which the first expression (5.85) is received with compressing force Ntaken according to its absolute value. To obtain the second law (5.85) (for partII in Fig. 5.14) we must replace in (H.3) N/2a, εx, σo by σo, εy, 0 respectively.

In the same manner the case of material flow between slightly inclinedplates can be considered (/10, 49/).

I

Appendix

Inelastic Zones Near Crack in Massifat Tension

We rewrite (5.99) asR = Fm+1/I1(m). (I.1)

Then at m = 1 we receive from (3.90)

R = (sin2 θ)/2π. (I.2)

While deriving (5.95) we neglect f′ in F. If we do not make it we haveinstead

f1v + (3m + 7)f ′′/(m + 1)2 + (2m + 1)(m + 2)f/(m + 1)4

+ 4m(m − 1)f ′′2/((2m + 1)f + (m + 1)2f ′′) = 0.

Taking as in the general case f(θ) proportional to eβθ we deduce linear algebraicequation

β6 + (4m2 + m + 8)β4/(m + 1)2 + (2m + 1)(4m + 9)β2/(m + 1)4

+ (2m + 1)2(m + 2)/(m + 1)6 = 0.(I.3)

At m = 3 e.g. we calculate

β6 + 2.9375β4 + 0.5742β2 + 0.06 = 0

and if we neglect the last small member we have above the solution β = 0 alsoβ2 = −0.23 and β2 = −2.727 (according to the Cardano’s relations the realsolution is −2.736), and so we can detail (5.96) as

f = 0.776 cos 0.46θ − 0.162 sin 0.46θ + 0.224 cos 1.65θ + 0.045 sin 1.65θ, (I.4)

compute I1(3) = −0.013 and according to (5.94), (I.1) and (I.4) – R-value.

210 I Appendix

Table I.1. Values of R(θ) at different m

m 1 3 15θ0 0 0.97 5 × 1017

π/4 0.08 30.9 1026

π/2 0.16 63.7 3.5 × 1028

3π/4 0.08 0.29 3 × 1027π 0 0.04 0.002

Similarly, for m = 15 we derive

β6 + 3.605β4 + 0.033β2 + 0.001 = 0.

Neglecting once again the last member we detail (5.96) as

f = 0.952 cos 0.0956θ−3.095 sin 0.0956θ+0.048 cos 1.8964θ+0.156 sin 1.8964θ,

compute I1(15) = −3.23 × 10−28 and, consequently, R. Its values for m = 1,3, 15 are given in Table I.1, and we can see from it that with a growth of mthe inelastic zone changes its form and dimensions immensely.

J

Appendix

Inelastic Zones Near Crack Endsat Transversal Shear

We rewrite (5.100) as

R(θ) = Fm+1/I(m)/f ′′′(π)/m (J.1)

whereR = 2r(2τe)m+1GΩ(t)/τ2(κ + 1)π.

At m = 1 we have from (J.1) relation

R = (1 + 3 cos2 θ)/2π

which coincides with equation for τe from (3.106)For m = 3 we have I(m) = 3.33 and

f = 0.997 cos 0.246θ + 0.555 sin 0.246θ − 0.997 cos 1.873θ + 0.461 sin 1.873θ.

Then we find f ′(θ), f ′′(θ) and /f ′′′(π)/.3 = 54.34. So, from (5.94) we have

F2 = (f ′′ + 0.4375f)2 + 2.25f ′2.

Table J.1. Values of R(θ) at different m

m 1 3 15θ0 0.64 1.10 0.21π/4 0.40 0.52 0.10π/2 0.32 0.41 0.183π/4 0.40 0.71 0.24π 0.64 1.14 0.35

212 J Appendix

For m = 15 according to boundary conditions we detail (5.96) as

f = −1.04 cos 0.162θ − 0.515 sin 0.162θ + 1.04 cos 2.146θ − 0.505 sin 2.146θ.

Then we compute f ′(θ), f ′′(θ) and /f ′′(π)/15 = 7.14×1010. So, from (5.94)we find

F2 = (f ′′ + 0.121f)2 + 3.52f ′2.

The values R at τe = 0.5 are given in the Table J.1 and we can see that allthe curves have similar form.

K

Appendix

Flow of Material in Cone

In order to demonstrate the computations in general case we make them forthe option µ = 1/3 when equation (5.119) becomes

dΘ/dχ = Θ/ tan χ − (4/3) tan 2ψ + (4 + Θ(1.5Θ − 5))(tan 2ψ)/3Ψ

− Θ(4 + 0.75Θ(tan 2ψ/ tan χ + 1 + 1/ sin2 χ)) sin 4ψ/3Ψ (K.1)

where according to (5.49)

3Ψ = 1 + 2 cos2 2ψ.

We integrate (K.1) at different Θ(0) = Θo and (5.118) at boundary con-dition ψ(0) = 0 by the finite differences method. The curves ψ(χ) at Θo = 1,2, 4 are marked in Fig. K.1 by numbers 1, 2, 3 respectively.

30

23

1

15

0 30 60 χ°

Ψ°

Fig. K.1. Diagrams ψ(χ)

214 K Appendix

Then we integrate the second expression (5.113) as

g = exp(

2

χ∫

0

(1/θ − 1.5) tan 2ψdχ)

neglecting an arbitrary multiplier because it is in numerator as well as indenominator of (5.120). Lastly we compute J3 and use (5.120). The diagramsof max τe(λ) is given in Fig. 5.22 by interrupted by points line and we can seethat it is near to that one at µ = 2/3.

L

Appendix

The Use of Hypotheses of Creep

We shall apply them to the case when (see sub-paragraph 1.5.5)

ε = Bσmtn (0 ≤ n ≤ 1). (L.1)

The validity of this expression at σ = constant represents a time hardeningtheory which is mainly used in this book. Differentiation of (L.1) atσ = constant gives the flow hypothesis

dε/dt = Bnσmtn−1. (L.2)

Combination of (L.1) and (L.2) allows to formulate strain hardening theoryin relation

ε1/n−1dε/dt = nB1/nσm/n. (L.3)

The well-known heredity integral must be written as following:

ε = Bn

1∫

0

σ(ξ)(t − ξ)n−1dξ. (L.4)

At σ = constant equations (L.2). . .(L.4) coincide with (L.1).To appreciate the validity of these hypotheses we must use them for the

regimes with σ = constant e.g. σ = ut where u = constant. In this case wehave a common expression

ε = KBσmtn (L.5)where K is equal to 1, n/(m + n), (n/(m + n))n, Γ(m+1)Γ(n+1)/Γ(m+n+1)for relations (L.1), (L.2), (L.3), (L.4) respectively. As for non-linear unsteadycreep m is big and n – small the flow hypothesis predicts negligible K-valuesagainst other theories, e.g. for m = 3, n = 1/8 we receive K = 0.04, 0.67and 0.8. The experiments /14/ give as a rule data between the predictionsof strain hardening and heredity hypotheses and in this situation a specialmeaning receives the time hardening theory with its advantage of simplicityfor practical applications.

216 L Appendix

To use the Hoff’s method we put in (L.2), (L.3) expression (1.42) andintegrate in limits 0, t∞ and 0, ∞ for t, ε that gives respectively

t∞ = (mB(σo)m)−1/n, t∞ = n1/nΓ(1/n + 1)(mB(σo)m)−1/n. (L.6)

At n = 1 (a steady creep) they coincide with (1.44). At n < 1 the first of themleads to bigger values of the fracture time than the second one.

Lastly, we compute the fracture and critical times according to strainhardening theory and (1.46) at α = 0, Ω(t) = Btn as

t∞/t∗ = (ne)1/nΓ(1/n + 1)

and we can see that at n = 1 value of t∞/t∗ is equal to e (the Neper’snumber) and with a fall of n this ratio increases.

M

Appendix

Use of the Coulomb’s Law for Description of SomeElastic-Plastic Systems at Cycling Loading

In some productions, for example, at deep boring the special device is used. Itconsists of the system of rings with conic polish sides (in Fig. M.1, a three suchrings are shown). The main difficulty of the use of similar devices is findingsuch loading regimes at which the rings are sliding relatively to each other.Here the results of experimental-theoretical study of the apparatus is made(it was fulfilled by the author at Leningrad Polytechnic).

The experiments were provided on the Amsler’s press and cycling machineCDM PU-100. At quasi-static loading the values of force P were taken from the

a) b)

N

T

q

x

A

P/πDα P

D/2 D/2

P

Fig. M.1. Scheme of ream of rings with their cross-section

218 M Appendix

P,MN

0.20

1

24

3

5

0.15

0.10

0.05

0 2 4 k m.10−3

Fig. M.2. Results of experiments on quasi-static loading

Table M.1. Data of quasi-static loading

No a b f C,MN/mMN/m

1 51.5 18.2 0.143 2872 35 30 0.024 2813 36 8 0.153 2984 3ß 7,5 0.136 2545 38 12.5 0.138 264

press scale and deformation κ – from two indicators. The simultaneous recordof P and κ at cycling loading was fulfilled by special electronic apparatus90-16.2. To make the tests in liquid the device was situated in a cylinder.

The results of tests at quasi-static loading at different environment andα are given in Fig. M.2 (solid lines 1 for air at α = 0.331, 2 – for lubricantwith graphite at α = 0.331, broken line 3 – for air at α = 0.262, interruptedby points lines 4 – for oil at α = 0.262 and 5 – for oil at α = 0.296) and inTable M.1 under the same numbers. The character of graphs P(κ) which havethe form of triangles with upper sides responding to loading and vertical aswell as lower straight lines – to unloading shows an opportunity of theoreticalprediction of deformation characteristics of the device.

At the increase of upper or lower rings diameter Db (Fig. M.1,a) on value∆Db the system becomes shorter on ∆κb = 0.5∆Db cot α. Similarly for otherrings we have ∆κs = ∆Ds cot α. As a result the whole shortening of the systemof n rings is

κ = (∆Db + (n − 2)∆Ds) cot α. (M.1)

M Appendix 219

When diameter D changes its value on ∆D the circumferential stressesappear. They can be replaced by radial forces q on the unit length of theperimeter (Fig. M.1, b) as

q = 2EA∆Di/Di2 (i = b, s) (M.2)

where E is modulus of elasticity, A – area of cross-section of a ring (shadedin Fig. M.1).

Making a sum of projections of forces on axis x in Fig. M.1, a we have forupper or lower and middle rings respectively

− (P/πDb) cos α + qb sin α + f(qb cos α + (P/πDb) sin α) = 0,− (P/πDs) cos α + 0.5qs sin α + f(0.5qs cos α + (P/πDs) sin α) = 0.

(M.3)

Here friction force T is found from the Coulomb’s law T = fN, f – coefficientof friction and normal component N is computed according to the equilibriumequation.

Replacing in (M.3) values qb, qs by relations (M.2) and putting values∆Db, ∆Ds into (M.1) we find after transformations

P = Cκ tan(α + ϕsign(dκ/dt)) tan α (M.4)

where sign (dκ/dt) reflects the difference of link between P, κ at loading andunloading, ϕ – angle of friction and C is a rigidity factor which at Ds = Db isgiven by expression

C = πEA/(n − 1.5)D. (M.5)

When a, b are tangents of angles between straight lines and axis κ inFig. M.2 at loading and unloading respectively, friction coefficient and rigidityfactor can be found as

f2 − 2(a + b)/(a − b) sin 2α + 1 = 0,

C2 − 0.5C(a + b)(cot2 α − 1) − ab cot2 α = 0.(M.6)

In the solutions we must take before square root sign minus for the firstrelation and plus – for the second one.

Values a, b are also given in Table M.1 and we can see that f-values arenear to their meanings in reference books and mean value of C = 277MN/mcoincides with that for tested reams at n = 7, E = 2x105 MN/m2, A =3.15x10−4 m2, D = 0.13m according to (M.5).

The analysis of the test results on quasi-static loading gives an opportunityto suppose two work types of the device at cycling loading – with mutual slipof rings and without it (as a piece of tube) with rigidity coefficients muchmore than follows from (M.4) as

C tan(α ± ϕ) tan α.

220 M Appendix

P,kN

40

20

5

812 4

39

1

2

6

10

11

A

B7

0 0.4 0.8 1.2 k,mx10−3

Fig. M.3. Function P(κ) at cycling loading

Table M.2. Data of cycling loading

No of loop in frequency P1 P2 ψ

regime figure in Hc kN test theory

I 1-4 21 25 5 2.86 2.70II 5-8 21 35 5 0.84 1.18III 9-12 21 55 5 0.87 0.72

0-A – B-0 statics any 0 0-32 0-32

The latter regime takes place if at the beginning of loading the minimum valueof force in a cycle is more than force P* responding to the beginning of ringsslip at unloading. For P* we have from (M.4)

P∗ = P1(tan(σ + ϕ))/ tan(σ − ϕ) (M.7)

where P1 is maximum load in a cycle.In Fig. M.3 as an example the test results for rings ream with α = 0.262

in oil at frequency 21 Hc in regimes I...III of Table M.2 together with brokenstraight lines of loading OA and unloading OB at quasi-static deformationare given. It is seen some difference of form and position of hysteresis loops atquasi-static and quick loadings. It is biggest for the regime with maximum ofamplitude that may be explained by the divergence from the Coulomb’s lawwith growth of a velocity of mutual displacement of sliding surfaces.

M Appendix 221

Now we find absorption coefficient ψ as the ratio of hysteresis loop areato potential energy 0.5Cκa of the system where κa is amplitude of the de-formation. Theoretical values of ψ are computed according to following from(M.4) relation

ψ = 8((P1 + P2)(1 + f2) tan α − (P1 − P2)f(1 + tan2 α))f sin α/((P1 − P2)×(1 + f2) tan α − (P1 + P2)f(1 + tan2 α))(1 − f2 tan2 α))×(1 − f2 tan2 α)) cos2 α.

where P2 is minimum force in a cycle.In conclusion we must notice that the difference between quick and slow

deformations of the ream is not big. Using the data on quasi-static loading wecan predict the character of the reams work including their dissipative lossesat cycling loading.

The received results may be used for explanation of behavior of otherelastic-plastic systems including some soils.

N

Appendix

Investigation of Gas Penetration in Polymersand Rubbers

Since polymers and rubbers are often used as shells and membranes (seeSects. 6.24, 6.2.5.) it is necessary to study gas penetration through them. Here-with such investigation is made (it was fulfilled by the author in LeningradState University).

The process of gas penetration in materials is often considered as successivephenomena of absorption and diffusion that go into direction of decreasingconcentration gradient.

Unsteady state of diffusion gas stream in a material is described by differ-ential equation similar to (1.12)

∂C/∂t = D∂2C/∂x2. (N.1)

Here D is diffusion coefficient, C – concentration of gas, x – coordinate in thedirection of its movement.

Steady state of diffusion stream subdues to relation

q = −D∂C/∂x (N.2)

where q – quantity of gas penetrating through unit surface area of a materialin unit time.

It is supposed that the gas absorption in polymers is described by theHenri’s law

C = βp. (N.3)

Here β – absorption coefficient, p – gas pressure.With consideration of (N.3) expression (N.2) may be represented in form

Q = Dβ∆pAt/h (N.4)

224 N Appendix

where Q is a volume of gas that is diffused through a sample of material withthickness h and area A during time t at difference of pressures before andafter the plate ∆p.

If we define the penetration factor r as gas volume which goes throughunit area af cross-section in second at unit gradient of pressure that is

r = Qh/At∆p (N.5)

constants r, D, β will be linked by relation

r = Dβ. (N.6)

Hence the gas penetration may be fully characterized by three constants – ofdiffusion D, absorption β and penetration r. They can be found experimentallyby different methods.

One of them allows to determine all the constants simultaneously from onetest as follows.

For the plate at initial and border conditions

C = Co at t = 0, 0 ≤ x ≤ h,

C = C1 at x = 0,C = C2 at x = h for all t

we have the following solution of (N.1)

C(x, t) = C1 + (C2 − C1)x/h + (2/π)∞∑

n=1

((C2 cos nn − C1)/n) sin(nπx/h)

× exp(−Dn2π2t/h2) + (4Co/π)∞∑

m=0

(2m + 1)−1 sin((2m + 1)πx/h)

× exp(−D(2m + 1)2π2t/d2). (N.7)

The gas stream through unit area of the plate cross-section into volumeVo is determined by expression

Vo∂C/∂t = −D∂C/∂x∣∣∣x=h

. (N.8)

Differentiating (N.7) by x and putting the result at x = h into (N.8) wereceive after integration from 0 to t following relation for gas volume Q whichwent through the plate at time t

Q = βA((p1 − p2)Dt/h + (h/π2)

(2

∞∑n=1

(p1 −p2 cos πn)(1− exp(−Dn2π2t/h2))

+4po

∞∑m=0

(cos(2m + 1)π)(1 − exp(−D(2m + 1)2π2t/h2))

)(N.9)

N Appendix 225

where po is the pressure at initial gas concentration Co in the plate, p1 andp2 – pressures before and after the plate.

At big t (N.9) becomes

Q = βA((p1 − p2)Dt/h − (p1 + 2p2)h/6 + poh/2). (N.10)

If Q = 0 we can find the “time of lagging” (at unsteady penetration) t1 as

t1 = h2((p1 + 2p2 − 3po)/6D(p1 − p2). (N.11)

At small p2 and po comparatively to p1 we receive

D = h2/6t1. (N.12)

So, when we have experimental dependence Q(t) for unsteady and steadystate of diffusion gas stream we can find with the help of relations (N.5),(N.12) and (N.6) constants r, D and β.

For the experiment a plant was created which gives an opportunity to testmaterial plates with thickness 0.1...1 cm and working area 110 cm2. Durationof a test is determined by the time when the diffusion velocity becomes con-stant and it is equaled usually to 2t1. The volume of a gas for computation isfound according to the Clapeyron’s relation

Q = 3.6HΣ∆Qi/T (N.13)

Q

30

1 3

2

6

5

4

20

10

050

t1100 150 200 250 t, h

cm3

Fig. N.1. Diagrams Q(t)

226 N Appendix

Table N.1. Data of tests

No material ∆p number of h rx1010 Dx107 βx103

cm2 samples cm cm4/sN cm2/s cm2/N

1 rubber 0.5 4 0.22 5.76 1.64 3.542 rubber 4 5 0.30 0.41 0.57 0.783 polyvinilchlorid 0.5 5 0.23 4.26 1.29 3.334 polypropilen 1 6 0.22 1.36 0.52 2.655 polyethilen 0.5 5 0.31 0.08 0.92 1.366 polycarbonat 2 5 0.27 0.47 0.22 1.93

where ∆Qi is the gas volume at considered time, H – atmospheric pressurein cm of a mercury pillar, T – temperature of in K. The diagrams Q(t) for6 materials are given in Fig. N.1 and computed according to relations (N.5),(N.12), (N.6) constants - in Table N.1.

The analysis of the results (here only a part of them is represented) showsthat the penetration of the materials falls with a growth of pressure difference.It is also seen that the fall of r takes place due to a decrease of D and β. Thechange of penetration should be taken into account at a construction of rubberand polymer structures.

Elsoufiev Sergiy Alexeevich, Vierhausstr.27,44807 Bochum [email protected], [email protected]

Curriculum Vitae

ELSOUFIEV Serguey (ELSUF’EV Sergiy)was born in November 1935 in Omsk (Siberia,Russia). From February 1997 lives with hiswife and a stepson constantly in Germany.Graduated with a distinction from theHydrotechnical Faculty of Leningrad Poly-technic Institute in 1959 (the diploma projectwas presented in English). Worked as anengineer at building sites. In 1964 received hismaster’s degree on the experimental bases ofthe Plasticity Theory. In 1985 defended hisdoctorate thesis in which he grounded a newscientific direction on strength computation ofstructures made of plastic metals, polymers,

rubbers, soils etc. On the theme published 135 works, 29 of them in Englishand 15 his works are re-edited in USA. The book “Geomechanics” is editedin Odessa State University.

From 1964 to 1970 worked as a scientist in the Polymer Strength Labora-tory at Leningrad State University, then as assistant, associate professor andprofessor in Leningrad Institutes (Polytechnic and Water transport). From1987 till 1997 - as a professor, the Head of the Strength of Materials Depart-ment in Odessa State Academy of Civil Engineering and Architecture. Deliv-ered lectures on the Strength of Materials and Structural Mechanics (3 yearsto foreign students in English), Theory of Elasticity, Plasticity and Creep,Applied Mechanics which combines very closely the Mechanism Theory, Ma-chines Parts and Strength of Materials, and additionally in Leningrad andOdessa Universities, as well as in special institute - the Fracture Mechanics -a unique and original discipline. All the courses above are published (partly inEnglish). Supervises the post-graduate students. For many years is a memberof master and doctorate Councils on different branches of Mechanics. Is amember of GAMM (Society for Applied Mathematics and Mechanics).

228 Curriculum Vitae

Speaks Russian, English, German, Polish, understands French, Ukranian.Has driving license. Healthy, is busy with sport (jogging, cycling, swimming).Neither drinks, nor smokes.

The contact address: Elsoufiev S., Vierhausstr.27, 44807 Bochum BRD(Germany).

Tel. 0049(0)2349586213. E-Mail: [email protected] and [email protected]

Index

absorption coefficient, 221, 223absolute rigidity, 37, 182active pressure, 22acumulations of solid particles, 1, 2angle of divergence, 105angle of internal friction, 17angle of repose, 22angle of rotation, 29anisotropic materials, 36anisotropic factors, 36antiplane problem, 40apex, 58, 140asymptotic approach, 66axisymmetric problem, 40, 44, 72, 149

base of structure, 6beam on elastic foundation, 2, 61bearing capacity, 94, 123, 139, 144, 154beck form, 67biggest main strain, 36boundary conditions, 40Boussinsq’s problem, 74brittle fractures, 11byharmonic equation, 43

capillaries in earth, 12choice of structures places, 1centroidal field, 98centroids, 18, 98circular hole, 63, 99circular pile, 123, 154circular punch, 75coefficient of friction, 22, 219coefficient of stability, 110

coherence pressure, 104coherent soil, 104cohesiveness, 2combination of equilibrium equations,

43, 45compatibility equations, 33, 42, 45complex variables, 47, 62compression of cylinder, 119, 179compulsory flow, 56condensation of soil, 7conformal transformations, 48consolidation degree, 13constant displacement in transversal

shear, 70constitutive laws, 34coreCoulomb’s law, 23, 217crack in tension, 65, 71, 102crack mechanics, 16crack propagation, 17, 51, 66, 126creep phenomenon, 24critical stress, 17cupillaries in soil, 12cycling loadings, 9, 217cycloids, 93cylindrical rigidity, 15cylindrical surface of slip, 114

damage parameter, 4, 26Darcy’s law, 11dangerous surface, 115decomposition of complex variables, 49deep shear, 110

230 Index

deflection of beam, 28, 61delta of amplitude, 145derivative of complex variables, 47deviators, 34, 35, 36differential static equations, 33disintegration of rocks, 1disperse soils, 10ductile materials, 25

eccentric compression, 26effective stress, 35effective strain, 35, 120elastic bed, 61elastic foundation, 62energy balance, 70engineering geology, 1equations of Caushy-Riemann, 47equivalent layer, 109equivalent stress, 36equivalent strain, 36expenditure of water, 13

factor of stability, 110failure of slope, 88feeble matrix, 36filtration consolidation, 11filtration of water, 12filtration pressure, 112finite strains, 33first boundary problem, 63ferst critical load, 60flow within cone, 121foundations, 1, 55, 129Fracture mechanics, 52fracture toughness, 51free energy, 127friction coefficient, 22, 219frozen earth, 1full elliptic integral, 77

gamma-function, 127gas penetration, 223generalization of Boussinesq’s solution,

149generalization of Flamant’s problem,

128generalized Hooke’s law, 34geological schemes and cross-sections, 1gradient of function, 85

Griffith’s relation, 67Gvozdev’s theorems, 3, 98

hardening, 17heap of sand, 85Hoff’s method, 161homogeneity of soil, 9Hooke’s law, 9hydraulic gradient, 12hydrodynamic pressure, 112hydrotechnical structures, 79hypotheses of geometrical character, 40

ideal plasticity, 35, 43imaginary part of complex variable, 48inclined crack in tension, 71incompressible body, 36independence of integral J, 149influence of stress state type, 36integral static equations, 54intensity factor, 51invariants of tensor, 32isotropy plane, 37

kinematic hypotheses, 144

laminated soils, 36landslide stability, 116Laplace’s operator, 45L’Hopital’s rule, 201leaned slopes, 116limit of brittle strengthload-bearing capacity, 123, 139, 154loading plane, 37lock, 144Lode’s parameters, 36longitudinal shear, 40longtime strength, 24loss of stability, 14

main axes, 96main stresses, 32maximum shearing stress, 33maximum strain, 36media mechanics, 31mean stress, 34membranes, 176method of corner points, 195mixed boundary problem, 40

Index 231

mixed fracture, 166mode of fracture, 21modulus of elasticity, 9modulus of shear, 34Mohr’s circle, 22moment of inertia, 15moment of torsion, 81

natural pressure, 6Neper’s number, 13, 216

orthotropic directions, 36

parallel plates, 57passive pressure, 22penetration of gas, 223penetration of wedge, 94, 139perfect plastic body, 86permeability, 2phases of soil state, 6place of prop structure, 1, 116, 117plane deformation for Hooke’s Law, 42plane shear, 110plane stress state, 42plastic hinges, 18plasticity theory methods, 3play-wood, 37Poisson’s ratio, 6polar coordinates, 41porosity, 2potential function, 146power law, 25Prandtl’s solution, 94primary creep, 24principle of Saint-Venant, 3propagate, 17propagation of crack, 146

rate of soil’s stabilisation, 10rays of fracture, 19real part of complex variable, 47reliability degree, 1repose angle, 22residial stress, 2resultant of soil pressure, 21, 108, 110retaining wall, 2, 22, 55rheological equations, 34Rice approach, 86

rigolith, 1rock, 1role of the bed, 62

second ultimate load, 108sediments, 1, 2semi-infinite solid, 74semi-plane under vertical load, 64semi-space, 74shearing displacements, 7sine of amplitude, 145skeleton’s volume, 8skid, 144sliding support, 144slip arc, 111slip element, 96slip lines, 7, 90, 93slopes stability, 2, 111slope under one-sided load, 131sluice, 144solid core, 7solution in series, 64specific coherence, 111specific gravity, 12specific weight, 6spherical coordinates, 45squizing water out, 10stability of footings, 2stable crack, 66statically determinate problem, 41, 43statically indeterminate problem, 33stick condition, 54stratified soil, 36strength condition for crack, 67stress intensity coefficient, 67stresses under rectangles, 75stretching energy, 51subtangent, 182sucker with holes, 11summation “layer by layer”, 78superposition method, 74surface of stretching, 17

tension under hydrostatic pressure, 161tensor of strains, 33tensor of stresses, 31Terzaghi-Gersewanov model, 11torus of revolution, 174transversally isotropic plate, 36

232 Index

transversal shear, 69, 148true strain, 25

ultimate plastic state theory, 17ultimate plasticity, 182unstable deformation, 4

volume of heap, 85

wedge pressed by inclined plates, 52wedge under concentrated force, 58wedge under one-sided load, 53

yielding demand, 117yielding point in shear, 35

Zhoukovski’s relation, 48

Foundations of Engineering Mechanics Series Editors: Vladimir I. Babitsky, Loughborough University, UK

Jens Wittenburg, Karlsruhe University, Germany

Alfutov, N.A. Stability of Elastic Structures, 2000 ISBN 3-540-65700-2

Astashev, V.K., Babitsky, V.I., Kolovsky, M.Z., Birkett, N. Dynamics and Control of Machines, 2000 ISBN 3-540-63722-2

Kolovsky, M.Z., Evgrafov, A.N., Semenov Y.A, Slousch, A.V. Advanced Theory of Mechanisms and Machines, 2000 ISBN 3-540-67168-4

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